链接
传送门
题意
给出一个\(n*n\)的矩阵，起点在左上角，每次只能向右或向下，求到到右下角的路径使其所有数字的乘积末尾的零的个数最少。输出零的个数和路径。
思路
末尾零的个数即为路径之积分解后2的指数和5的指数的最小值，首先预处理出所有数字分解后2和5的指数，然后类似于数字三角形的DP。唯一的trick是数字中出现0。
代码
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| #include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int inf = 50000;
const int maxn = 1010;
const int fac[] = {2, 5};
int a[maxn][maxn][2];
int d[maxn][maxn][2];
char path[maxn][maxn][2], s[maxn << 1];
int main() {
int n;
scanf("%d", &n);
bool flag = false;
int posx = 0, posy = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
int x;
scanf("%d", &x);
if (x == 0) {
a[i][j][0] = a[i][j][1] = inf;
posx = i, posy = j;
flag = true;
continue;
}
for (int k = 0; k < 2; ++k) {
while (x % fac[k] == 0) {
x /= fac[k];
++a[i][j][k];
}
}
}
}
for (int i = 1; i <= n; ++i) {
d[0][i][0] = d[i][0][0] = d[0][i][1] = d[i][0][1] = inf;
}
d[1][1][0] = a[1][1][0], d[1][1][1] = a[1][1][1];
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
if (i == 1 && j == 1) continue;
for (int k = 0; k < 2; ++k) {
if (d[i - 1][j][k] > d[i][j - 1][k]) {
d[i][j][k] = d[i][j - 1][k] + a[i][j][k];
path[i][j][k] = 'R';
}
else {
d[i][j][k] = d[i - 1][j][k] + a[i][j][k];
path[i][j][k] = 'D';
}
}
}
}
int ans = d[n][n][0] > d[n][n][1] ? 1 : 0;
if (flag && d[n][n][ans] >= 1) {
puts("1");
for (int i = 1; i < posx; ++i) {
putchar('D');
}
for (int i = 1; i < posy; ++i) {
putchar('R');
}
for (int i = posx; i < n; ++i) {
putchar('D');
}
for (int i = posy; i < n; ++i) {
putchar('R');
}
puts("");
return 0;
}
printf("%dn", d[n][n][ans]);
int len = (n - 1) << 1, x = n, y = n;
while (len) {
s[--len] = path[x][y][ans];
if (s[len] == 'D') {
--x;
}
else {
--y;
}
}
puts(s);
return 0;
}
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