feat: 2245. Maximum Trailing Zeros in a Cornered Path : 前缀和

Author: upupming

leetcode/contest-289/6072. Maximum Trailing Zeros in a Cornered Path.js

@@ -2,77 +2,42 @@
  * @param {number[][]} grid
  * @return {number}
  */
-const maxTrailingZeros = function (g) {
-  const m = g.length
-  const n = g[0].length
-  // { c2: 前面的数, c5:, c10: 多少 0 }
-  const el = (c) => {
-    const com = Math.min(c.c2, c.c5)
-    c.c2 -= com
-    c.c5 -= com
-    c.c10 += com
+const maxTrailingZeros = function (grid) {
+  const m = grid.length; const n = grid[0].length
+  let ans = 0; const base = 1e7
+  const sr = []; const sc = []
+  const cnt = (a) => {
+    let c2 = 0; let c5 = 0
+    while (a % 2 === 0) a /= 2, c2++
+    while (a % 5 === 0) a /= 5, c5++
+    return [c2, c5]
   }
-  const mul = (a, b) => {
-    const c = {
-      c2: a.c2 + b.c2,
-      c5: a.c5 + b.c5,
-      c10: a.c10 + b.c10
-    }
-    el(c)
-    return c
-  }
-  const div = (a, b) => {
-    const c = {
-      c2: a.c2 - b.c2,
-      c5: a.c5 - b.c5,
-      c10: a.c10 - b.c10
-    }
-    el(c)
-    return c
-  }
-  const get = (c) => {
-    cc = { c2: 0, c5: 0, c10: 0 }
-    while (c % 2 === 0) {
-      cc.c2++
-      c /= 2
-    }
-    while (c % 5 === 0) {
-      cc.c5++
-      c /= 5
-    }
-    el(c)
-    return cc
+  const get = (i, j, k) => {
+    return cnt(grid[i][j])[k]
   }
-  const sumR = []; const sumC = []
-  for (let i = 0; i < m; i++) {
-    sumR[i] = []
-    sumR[i][-1] = get(1)
-    for (let j = 0; j < n; j++) {
-      sumR[i][j] = mul(sumR[i][j - 1], get(g[i][j]))
-      // console.log(0, i, j, get(g[i][j]), sumR[i][j-1], sumR[i][j])
+  for (let j = 1; j <= n; j++) {
+    sr[j] = [0]
+    for (let i = 1; i <= m; i++) {
+      sr[j][i] = sr[j][i - 1] + get(i - 1, j - 1, 0) * base + get(i - 1, j - 1, 1)
     }
   }
-  for (let j = 0; j < n; j++) {
-    sumC[j] = []
-    sumC[j][-1] = get(1)
-    for (let i = 0; i < m; i++) {
-      sumC[j][i] = mul(sumC[j][i - 1], get(g[i][j]))
-      // console.log('1', j, i, sumR[i][j])
+  for (let i = 1; i <= m; i++) {
+    sc[i] = [0]
+    for (let j = 1; j <= n; j++) {
+      sc[i][j] = sc[i][j - 1] + get(i - 1, j - 1, 0) * base + get(i - 1, j - 1, 1)
     }
   }
-
-  let ans = 0
-  for (let i = 0; i < m; i++) {
-    for (let j = 0; j < n; j++) {
-      // 列 [0, i]，行 [j+1, n-1)
-      const tmp = mul(div(sumR[i][n - 1], sumR[i][j]), sumC[j][i])
-      // 列 [0, i]，行 [0, j-1]
-      const tmp1 = mul(sumR[i][j - 1], sumC[j][i])
-      // 列 [i, m-1]，行 [0, j-1]
-      const tmp2 = mul(sumR[i][j - 1], div(sumC[j][m - 1], sumC[j][i - 1]))
-      // 列 [i, m-1]，行 [j+1, n-1)
-      const tmp3 = mul(div(sumR[i][n - 1], sumR[i][j]), div(sumC[j][m - 1], sumC[j][i - 1]))
-      ans = Math.max(ans, tmp.c10, tmp1.c10, tmp2.c10, tmp3.c10)
+  // 枚举转角点
+  for (let i = 1; i <= m; i++) {
+    for (let j = 1; j <= n; j++) {
+      for (const [rs, re] of [[1, i], [i, m]]) {
+        // 避免重复计算 [i, j] 这个点
+        for (const [cs, ce] of [[1, j - 1], [j + 1, n]]) {
+          const a = (Math.floor(sr[j][re] / base) - Math.floor(sr[j][rs - 1] / base)) + (sc[i][ce] / base - sc[i][cs - 1] / base)
+          const b = (Math.floor(sr[j][re] % base) - Math.floor(sr[j][rs - 1] % base)) + (sc[i][ce] % base - sc[i][cs - 1] % base)
+          ans = Math.max(ans, Math.min(a, b))
+        }
+      }
     }
   }
   return ans