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update: 添加问题“3337.字符串转换后的长度II”的代码和题解(#936)
* 3337: WA.cpp (#935) Signed-off-by: LetMeFly666 <[email protected]> * archive: before changing computer (#935) 还是先全提交到这个分支上吧，这道题还是尽量做完然后合并到master上比较好 Signed-off-by: LetMeFly666 <[email protected]> * 3337: WA.cpp (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: WA.cpp (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: WA.cpp (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - WA (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - Ok (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: WA.cpp (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - WA (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - AC (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - WA (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - AC (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: test.cpp - WA (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: AC.cpp (#935) Signed-off-by: LetMeFly666 <[email protected]> * docs: 预写题解 (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: WA.py (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: AC.py - AC,6.25%,71.25% (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: AC.cpp - AC,40.84%,94.37% (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: CEjava (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: CEjava (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: CEjava (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: RE.java (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: TLE.java (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: AC.java - AC,17.24%,82.76% (#935) F**k 调了好久 Signed-off-by: LetMeFly666 <[email protected]> * words: 昨日、前天 英语单词 (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: Init.go (#935) Signed-off-by: LetMeFly666 <[email protected]> * 3337: AC.go - AC,26.19%,95.24% (#935) 直接打一遍过喽 Signed-off-by: LetMeFly666 <[email protected]> * docs: rename (#935) Signed-off-by: LetMeFly666 <[email protected]> * update: 添加问题“3337.字符串转换后的长度II”的代码和题解(#936) Signed-off-by: LetMeFly666 <[email protected]> --------- Signed-off-by: LetMeFly666 <[email protected]>
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‎.gitlog‎

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‎Codes/3337-total-characters-in-string-after-transformations-ii.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-05-14 09:36:25
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-05-14 23:47:59
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* @Description: AC,40.84%,94.37%
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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typedef long long ll;
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typedef array<array<ll, 26>, 26> Matrix;
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/*
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1 1 1 1 ... 2
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*/
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class Solution {
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private:
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static const int MOD = 1000000007;
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Matrix Pow(Matrix a, int b) {
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Matrix ans{};
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for (int i = 0; i < 26; i++) {
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ans[i][i] = 1;
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}
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while (b) {
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if (b & 1) {
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ans = Mul(ans, a);
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}
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a = Mul(a, a);
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b >>= 1;
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}
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return ans;
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}
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Matrix Mul(Matrix& a, Matrix& b) {
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Matrix ans{};
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for (int i = 0; i < 26; i++) {
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for (int j = 0; j < 26; j++) {
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for (int k = 0; k < 26; k++) {
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ans[i][k] = (ans[i][k] + a[i][j] * b[j][k] % MOD) % MOD;
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}
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}
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}
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return ans;
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}
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public:
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int lengthAfterTransformations(string s, int t, vector<int>& nums) {
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Matrix M{};
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for (int i = 0; i < 26; i++) {
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for (int j = 1; j <= nums[i]; j++) {
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M[i][(i + j) % 26] = 1;
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}
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}
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M = Pow(M, t);
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ll cnt[26] = {0};
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for (char c : s) {
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cnt[c - 'a']++;
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}
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int ans = 0;
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for (int i = 0; i < 26; i++) {
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for (int j = 0; j < 26; j++) {
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ans = (ans + M[i][j] * cnt[i] % MOD) % MOD;
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}
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}
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return ans;
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}
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};

‎Codes/3337-total-characters-in-string-after-transformations-ii.go‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-05-15 09:49:53
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-05-15 10:05:13
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*/
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package main
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var MOD3337 = int64(1000000007)
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type matrix3337 [26][26]int64
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func pow(a matrix3337, b int) (ans matrix3337) {
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for i := 0; i < 26; i++ {
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ans[i][i] = 1
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}
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for ; b > 0; b >>= 1 {
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if b & 1 == 1 {
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ans = mul(ans, a)
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}
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a = mul(a, a)
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}
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return
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}
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func mul(a, b matrix3337) (ans matrix3337) {
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for i := 0; i < 26; i++ {
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for k := 0; k < 26; k++ {
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for j := 0; j < 26; j++ {
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ans[i][j] = (ans[i][j] + a[i][k] * b[k][j]) % MOD3337
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}
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}
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}
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return
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}
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func lengthAfterTransformations(s string, t int, nums []int) int {
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M := matrix3337{}
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for i, d := range nums {
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for j := 1; j <= d; j++ {
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M[i][(i + j) % 26] = 1
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}
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}
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Mt := pow(M, t)
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times := make([]int64, 26)
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for i := 0; i < len(s); i++ {
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times[s[i] - 'a']++
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}
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ans := int64(0)
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for i := 0; i < 26; i++ {
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sum := int64(0)
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for j := 0; j < 26; j++ {
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sum += Mt[i][j]
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}
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ans = (ans + sum * times[i]) % MOD3337
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}
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return int(ans)
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}

‎Codes/3337-total-characters-in-string-after-transformations-ii.java‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-05-14 22:01:41
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-05-15 00:12:57
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*/
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import java.util.List;
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class Solution {
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private final int MOD = 1000000007;
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private long[][] pow(long[][] a, int b) {
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long[][] ans = new long[26][26];
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for (int i = 0; i < 26; i++) {
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ans[i][i] = 1;
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}
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while (b > 0) {
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if ((b & 1) == 1) {
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ans = mul(ans, a);
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}
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a = mul(a, a);
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b >>= 1;
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}
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return ans;
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}
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private long[][] mul(long[][]a, long[][] b) {
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long[][] ans = new long[26][26];
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for (int i = 0; i < 26; i++) {
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for (int k = 0; k < 26; k++) {
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for (int j = 0; j < 26; j++) {
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ans[i][j] = (ans[i][j] + a[i][k] * b[k][j] % MOD) % MOD;
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}
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}
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}
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return ans;
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}
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public int lengthAfterTransformations(String s, int t, List<Integer> nums) {
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long[][] M = new long[26][26];
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for (int i = 0; i < 26; i++) {
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for (int j = 1; j <= nums.get(i); j++) {
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M[i][(i + j) % 26] = 1;
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}
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}
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long[][] Mt = pow(M, t);
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long[] cnt = new long[26];
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for (int i = 0; i < s.length(); i++) {
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cnt[s.charAt(i) - 'a']++;
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}
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long ans = 0;
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for (int i = 0; i < 26; i++) {
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long sum = 0;
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for (int j = 0; j < 26; j++) {
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sum += Mt[i][j];
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}
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sum %= MOD;
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ans = (ans + sum * cnt[i]) % MOD;
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}
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return (int)ans;
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}
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}

‎Codes/3337-total-characters-in-string-after-transformations-ii.py‎

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'''
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Author: LetMeFly
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Date: 2025-05-14 22:01:41
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-05-14 23:46:10
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'''
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from typing import List
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MOD = 1000000007
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class Solution:
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def mul(self, a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
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ans = [[0] * 26 for _ in range(26)]
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for i in range(26):
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for k in range(26):
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for j in range(26):
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ans[i][j] = (ans[i][j] + a[i][k] * b[k][j]) % MOD
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return ans
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def pow(self, a: List[List[int]], b: int) -> List[List[int]]:
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ans = [[0] * 26 for _ in range(26)]
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for i in range(26):
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ans[i][i] = 1
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while b:
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if b & 1:
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ans = self.mul(ans, a)
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a = self.mul(a, a)
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b >>= 1
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return ans
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def lengthAfterTransformations(self, s: str, t: int, nums: List[int]) -> int:
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M = [[0] * 26 for _ in range(26)]
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for i, v in enumerate(nums):
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for j in range(1, v + 1):
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M[i][(i + j) % 26] = 1
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Mt = self.pow(M, t)
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cnt = [0] * 26
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for c in s:
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cnt[ord(c) - ord('a')] += 1
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ans = 0
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for i in range(26):
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ans += sum(Mt[i]) * cnt[i]
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return ans % MOD

‎Codes/3337-total-characters-in-string-after-transformations-ii.test.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-05-14 22:29:09
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-05-14 22:46:21
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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// #include "3337-total-characters-in-string-after-transformations-ii.cpp
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#endif
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typedef long long ll;
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// Copy And Change From 0x3f的题解
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class Solution {
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static constexpr int MOD = 1'000'000'007;
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static constexpr int SIZE = 26;
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using Matrix = array<array<ll, SIZE>, SIZE>;
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// 返回矩阵 a 和矩阵 b 相乘的结果
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Matrix mul(Matrix& a, Matrix& b) {
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Matrix c{};
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for (int i = 0; i < SIZE; i++) {
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for (int k = 0; k < SIZE; k++) {
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if (a[i][k] == 0) {
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continue;
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}
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for (int j = 0; j < SIZE; j++) {
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c[i][j] = (c[i][j] + (long long) a[i][k] * b[k][j]) % MOD;
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}
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}
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}
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return c;
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}
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// 返回 n 个矩阵 a 相乘的结果
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Matrix Pow(Matrix a, int n) {
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Matrix res = {};
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for (int i = 0; i < SIZE; i++) {
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res[i][i] = 1; // 单位矩阵
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}
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while (n) {
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if (n & 1) {
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res = mul(res, a);
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}
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a = mul(a, a);
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n >>= 1;
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}
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return res;
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}
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public:
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int lengthAfterTransformations(string s, int t, vector<int>& nums) {
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Matrix M{};
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for (int i = 0; i < SIZE; i++) {
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for (int j = i + 1; j <= i + nums[i]; j++) {
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M[i][j % SIZE] = 1;
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}
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}
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M = Pow(M, t);
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int cnt[SIZE]{};
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for (char c : s) {
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cnt[c - 'a']++;
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}
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int ans = 0;
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for (int i = 0; i < 26; i++) {
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for (int j = 0; j < 26; j++) {
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ans = (ans + M[i][j] * cnt[i] % MOD) % MOD;
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}
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}
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return ans;
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}
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};

‎README.md‎

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|3305.元音辅音字符串计数I|中等|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/12/LeetCode%203305.%E5%85%83%E9%9F%B3%E8%BE%85%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%AE%A1%E6%95%B0I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146197747" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-i/solutions/3607571/letmefly-3305yuan-yin-fu-yin-zi-fu-chuan-ptkg/" target="_blank">LeetCode题解</a>|
965965
|3306.元音辅音字符串计数II|中等|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/13/LeetCode%203306.%E5%85%83%E9%9F%B3%E8%BE%85%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%AE%A1%E6%95%B0II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146228751" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/solutions/3609545/letmefly-3306yuan-yin-fu-yin-zi-fu-chuan-7y2q/" target="_blank">LeetCode题解</a>|
966966
|3335.字符串转换后的长度I|中等|<a href="https://leetcode.cn/problems/total-characters-in-string-after-transformations-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/13/LeetCode%203335.%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%BD%AC%E6%8D%A2%E5%90%8E%E7%9A%84%E9%95%BF%E5%BA%A6I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147916305" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/total-characters-in-string-after-transformations-i/solutions/3675555/letmefly-3335zi-fu-chuan-zhuan-huan-hou-9ck9y/" target="_blank">LeetCode题解</a>|
967+
|3337.字符串转换后的长度II|困难|<a href="https://leetcode.cn/problems/total-characters-in-string-after-transformations-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/15/LeetCode%203337.%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%BD%AC%E6%8D%A2%E5%90%8E%E7%9A%84%E9%95%BF%E5%BA%A6II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147976391" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/total-characters-in-string-after-transformations-ii/solutions/3677448/letmefly-3337zi-fu-chuan-zhuan-huan-hou-h56wo/" target="_blank">LeetCode题解</a>|
967968
|3340.检查平衡字符串|简单|<a href="https://leetcode.cn/problems/check-balanced-string/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/14/LeetCode%203340.%E6%A3%80%E6%9F%A5%E5%B9%B3%E8%A1%A1%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146249653" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/check-balanced-string/solutions/3610899/letmefly-3340jian-cha-ping-heng-zi-fu-ch-p8eo/" target="_blank">LeetCode题解</a>|
968969
|3341.到达最后一个房间的最少时间I|中等|<a href="https://leetcode.cn/problems/find-minimum-time-to-reach-last-room-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/07/LeetCode%203341.%E5%88%B0%E8%BE%BE%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E6%88%BF%E9%97%B4%E7%9A%84%E6%9C%80%E5%B0%91%E6%97%B6%E9%97%B4I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147807022" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-minimum-time-to-reach-last-room-i/solutions/3672076/letmefly-3341dao-da-zui-hou-yi-ge-fang-j-3m6o/" target="_blank">LeetCode题解</a>|
969970
|3342.到达最后一个房间的最少时间II|中等|<a href="https://leetcode.cn/problems/find-minimum-time-to-reach-last-room-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/09/LeetCode%203342.%E5%88%B0%E8%BE%BE%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E6%88%BF%E9%97%B4%E7%9A%84%E6%9C%80%E5%B0%91%E6%97%B6%E9%97%B4II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147835524" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-minimum-time-to-reach-last-room-ii/solutions/3672695/letmefly-3342dao-da-zui-hou-yi-ge-fang-j-qrh6/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 3335.字符串转换后的长度I.md‎

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>
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> 接着从`y`到`a`遍历，将cnt[i]赋值给cnt[i + 1]；
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>
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> 最后令`cnt[0] := z`，`cnt[1] += z`，`ans += z`（这是因为没轮操作`z`会变成`ab`，同时字符串长度会加一）
106+
> 最后令`cnt[0] := z`，`cnt[1] += z`，`ans += z`（这是因为每轮操作`z`会变成`ab`，同时字符串长度会加一）
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+ 时间复杂度$O(len(s)\cdot C)$，其中$C=26$
109109
+ 空间复杂度$O(C)$

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