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灵茶の试炼 * 13
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‎codeforces-div1/src/main/java/p1179/CF1179B.java‎

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package p1179;
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import java.nio.charset.StandardCharsets;
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import java.util.ArrayList;
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import java.util.List;
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import java.util.Scanner;
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public class CF1179B {
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in, StandardCharsets.UTF_8);
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int n = scanner.nextInt();
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int m = scanner.nextInt();
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System.out.println(solve(n, m));
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}
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private static String solve(int n, int m) {
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List<String> resList = new ArrayList<>();
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int x1 = 1, y1 = 1, x2 = n, y2 = m;
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boolean flag = true;
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for (int i = 0; i < n * m; i++) {
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resList.add(x1 + " " + y1);
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if (flag) {
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if (y1 < m) {
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y1++;
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} else {
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x1++;
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}
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} else {
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if (y1 > 1) {
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y1--;
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} else {
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x1++;
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}
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}
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i++;
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if (i == n * m) {
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break;
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}
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resList.add(x2 + " " + y2);
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if (!flag) {
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if (y2 < m) {
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y2++;
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} else {
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x2--;
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flag = true;
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}
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} else {
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if (y2 > 1) {
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y2--;
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} else {
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x2--;
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flag = false;
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}
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}
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}
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return String.join(System.lineSeparator(), resList);
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}
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}
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/*
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B. Tolik and His Uncle
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https://codeforces.com/contest/1179/problem/B
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灵茶の试炼 2023-02-23
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题目大意：
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输入 n m (n*m≤1e6)，表示一个 n 行 m 列的棋盘，行列编号从 1 开始。
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初始时，你在 (1,1)。每一步，你可以使用一个方向向量 (dx,dy)，然后从当前位置 (x,y) 移动到 (x+dx,y+dy)。
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你需要访问每个格子恰好一次，且每一步使用的方向向量互不相同。
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如果存在这样的移动方案，输出任意一组符合要求的 n*m 个坐标，表示你每一步所在的位置。否则输出 -1。
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constructive algorithms
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https://codeforces.com/contest/1179/submission/108111481
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提示 1：方案一定存在。
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提示 2：从左上角跳到右下角，这个方向向量以后绝不会再用到了。
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提示 3：然后再从右下角跳到 (1,2)，这个方向向量以后也绝不会再用到了。
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======
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input
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2 3
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output
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1 1
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1 3
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1 2
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2 2
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2 3
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2 1
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input
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1 1
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output
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1 1
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*/

‎codeforces-div1/src/main/java/p388/CF388B.java‎

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package p388;
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import java.nio.charset.StandardCharsets;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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import java.util.Scanner;
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public class CF388B {
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in, StandardCharsets.UTF_8);
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int k = scanner.nextInt();
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System.out.println(solve(k));
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}
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// 2^0 ~ 2^29
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private static String solve(int num) {
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// 1 ~ 92
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char[][] adj = new char[93][93];
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for (char[] row : adj) {
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Arrays.fill(row, 'N');
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}
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for (int k = 0; k < 30; k++) {
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// a1 a - a'
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// / | x |
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// b1 - b - b'
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// ? ?
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// c1 c - c'
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int a = 3 * k + 3, b = a + 1, c = b + 1;
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int a1 = 3 * k, b1 = a1 + 1, c1 = b1 + 1;
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// b1-a, b1-b, a-b
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adj[b1][a] = adj[a][b1] = 'Y';
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adj[b1][b] = adj[b][b1] = 'Y';
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adj[a][b] = adj[b][a] = 'Y';
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// b 连 c 相当于加上 2^k
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if ((num >> k & 1) == 1) {
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adj[b][c] = adj[c][b] = 'Y';
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}
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if (k >= 1) {
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// a1-a, a1-b, c1-c
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adj[a1][a] = adj[a][a1] = 'Y';
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adj[a1][b] = adj[b][a1] = 'Y';
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adj[c1][c] = adj[c][c1] = 'Y';
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}
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}
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// 2-92
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adj[2][92] = adj[92][2] = 'Y';
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List<String> resList = new ArrayList<>();
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resList.add("92");
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// Consider the graph vertexes are numbered from 1 to n.
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for (int i = 1; i < 93; i++) {
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resList.add(new String(adj[i]).substring(1));
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}
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return String.join(System.lineSeparator(), resList);
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}
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}
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/*
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B. Fox and Minimal path
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https://codeforces.com/contest/388/problem/B
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灵茶の试炼 2023-02-28
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输入 k(1≤k≤1e9)。
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构造一个节点个数不超过 1000 的简单无向图（节点编号从 1 开始），使得从节点 1 到节点 2 的最短路径的数量恰好为 k。
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输出 n 以及一个 n*n 的邻接矩阵 g，如果 i 和 j 之间有边，则 g[i][j]='Y'，否则为 'N'。
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注意不能有自环，即 g[i][i] 必须为 'N'。
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constructive algorithms
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二进制构造。
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k = x1 * 2^0 + x2 * 2^1 + ... + xn * 2^n
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======
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input
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2
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output
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4
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NNYY
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NNYY
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YYNN
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YYNN
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input
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9
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output
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8
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NNYYYNNN
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NNNNNYYY
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YNNNNYYY
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YNNNNYYY
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YNNNNYYY
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NYYYYNNN
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NYYYYNNN
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NYYYYNNN
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input
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1
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output
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2
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NY
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YN
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*/

‎codeforces-div1/src/main/java/p708/CF708B.java‎

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package p708;
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import java.nio.charset.StandardCharsets;
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import java.util.Scanner;
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public class CF708B {
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private static final String IMPOSSIBLE = "Impossible";
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in, StandardCharsets.UTF_8);
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int a00 = scanner.nextInt();
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int a01 = scanner.nextInt();
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int a10 = scanner.nextInt();
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int a11 = scanner.nextInt();
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System.out.println(solve(a00, a01, a10, a11));
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}
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private static String solve(int a00, int a01, int a10, int a11) {
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// 根据 f(0,0) 可以求出 0 的个数 c0，f(0,0) = C(c0, 2)
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int cnt0 = f(a00);
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int cnt1 = f(a11);
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if (cnt0 < 0 || cnt1 < 0) {
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return IMPOSSIBLE;
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}
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if (a01 == 0 && a10 == 0) {
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if (a00 > 0 && a11 > 0) {
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return IMPOSSIBLE;
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} else if (a11 == 0) {
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return "0".repeat(cnt0);
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} else {
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return "1".repeat(cnt1);
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}
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}
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if (a01 + a10 != cnt0 * cnt1) {
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return IMPOSSIBLE;
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}
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int left1 = a10 / cnt0;
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int right0 = a10 % cnt0;
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String res = "1".repeat(left1) + "0".repeat(cnt0 - right0);
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if (right0 > 0) {
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res += "1" + "0".repeat(right0);
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cnt1--;
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}
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res += "1".repeat(cnt1 - left1);
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return res;
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}
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// a·x^2 + b·x + c = 0 一元二次方程求根公式 (-b ± sqrt(b^2 - 4ac)) / 2a
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// x(x-1)/2 = a
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// x = (1 + sqrt(1 + 8a)) / 2
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private static int f(long a) {
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int x = (int) (1 + Math.sqrt(a * 8 + 1.0) / 2);
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if (x * (x - 1L) / 2 == a) {
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return x;
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}
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return -1;
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}
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}
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/*
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B. Recover the String
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https://codeforces.com/contest/708/problem/B
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灵茶の试炼 2023-02-24
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题目大意：
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对于 01 字符串 s，定义 f(x,y) 表示子序列 [x,y] 在 s 中的出现次数。
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输入 f(0,0), f(0,1), f(1,0) 和 f(1,1)，范围在 [0,1e9]。
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请构造任意一个满足输入的非空字符串 s。
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如果不存在，输出 Impossible。
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注：子序列是从 s 中删除某些元素得到的。
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constructive algorithms
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https://codeforces.com/contest/708/submission/194669010
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提示 1：根据 f(0,0) 可以求出 0 的个数 c0，因为 f(0,0) = C(c0,2)；同理可求出 1 的个数 c1。
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提示 2：f(0,1) + f(1,0) = c0 * c1
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======
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input
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1 2 3 4
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output
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Impossible
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input
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1 2 2 1
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output
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0110
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*/

‎codeforces-div1/src/test/java/p1179/CF1179BTests.java‎

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package p1179;
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import base.AbstractOjTests;
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import org.junit.jupiter.api.Test;
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import java.io.IOException;
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public class CF1179BTests extends AbstractOjTests {
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public CF1179BTests() {
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super("/p1179/B/");
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}
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@Test
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public void example1() throws IOException {
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super.doSetSystemInOut(INPUT1);
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CF1179B.main(null);
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// 答案不唯一
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// super.doAssertion(OUTPUT1);
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}
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@Test
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public void example2() throws IOException {
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super.doSetSystemInOut(INPUT2);
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CF1179B.main(null);
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super.doAssertion(OUTPUT2);
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}
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}

‎codeforces-div1/src/test/java/p388/CF388BTests.java‎

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package p388;
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import base.AbstractOjTests;
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import org.junit.jupiter.api.Test;
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import java.io.IOException;
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public class CF388BTests extends AbstractOjTests {
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public CF388BTests() {
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super("/p388/B/");
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}
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@Test
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public void example1() throws IOException {
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super.doSetSystemInOut(INPUT1);
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CF388B.main(null);
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// 答案不唯一
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// super.doAssertion(OUTPUT1);
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}
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@Test
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public void example2() throws IOException {
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super.doSetSystemInOut(INPUT2);
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CF388B.main(null);
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// 答案不唯一
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// super.doAssertion(OUTPUT2);
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}
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@Test
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public void example3() throws IOException {
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super.doSetSystemInOut(INPUT3);
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CF388B.main(null);
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// 答案不唯一
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// super.doAssertion(OUTPUT3);
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}
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}

‎codeforces-div1/src/test/java/p708/CF708BTests.java‎

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package p708;
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import base.AbstractOjTests;
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import org.junit.jupiter.api.Test;
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import java.io.IOException;
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public class CF708BTests extends AbstractOjTests {
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public CF708BTests() {
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super("/p708/B/");
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}
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@Test
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public void example1() throws IOException {
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super.doSetSystemInOut(INPUT1);
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CF708B.main(null);
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super.doAssertion(OUTPUT1);
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}
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@Test
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public void example2() throws IOException {
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super.doSetSystemInOut(INPUT2);
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CF708B.main(null);
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// 答案不唯一
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// super.doAssertion(OUTPUT2);
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}
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}

‎codeforces-div1/src/test/resources/p1179/B/input1.txt‎

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2 3

‎codeforces-div1/src/test/resources/p1179/B/input2.txt‎

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1 1

‎codeforces-div1/src/test/resources/p1179/B/output1.txt‎

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1 1
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1 3
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1 2
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2 2
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2 3
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2 1

‎codeforces-div1/src/test/resources/p1179/B/output2.txt‎

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1 1

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