Codeforces #2B B. The least round way

#include <iostream>
第四题：这题求一条路径上的数，使得这些数相乘后后边的0的个数最少分析：这题一看就是DP问题，但是需要数学的一点分析，当时我一看就吓到了（数学弱渣T_T），然后去做E题了
对于表格中有0的情况，答案最多为1，路径便是通过这个点的路径；或者为0；咱们先抛弃这种情况不谈，把0当作10处理

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#define N 1007
using namespace std;
int g[2]

;
int dp[2]

;
int zx,zy;
int mat[2]

;
char path[4*N];
int n;
int solve(int mk){
int i,j;
g[mk][1][1] = 0;
dp[mk][1][1] = mat[mk][1][1];
for (i = 1; i <= n; ++i){
for (j = 1; j <= n; ++j){
if (i == 1 && j == 1) continue;
if (i == 1){
dp[mk][i][j] = dp[mk][i][j - 1] + mat[mk][i][j];
g[mk][i][j] = 1;
}
else if (j == 1){
dp[mk][i][j] = dp[mk][i - 1][j] + mat[mk][i][j];
g[mk][i][j] = 0;
}
else{
int tp1 = dp[mk][i - 1][j];
int tp2 = dp[mk][i][j - 1];
if (tp1 < tp2){
dp[mk][i][j] = dp[mk][i - 1][j] + mat[mk][i][j];
g[mk][i][j] = 0;
}
else{
dp[mk][i][j] = dp[mk][i][j - 1] + mat[mk][i][j];
g[mk][i][j] = 1;
}
}
}
}
return dp[mk]

;
}
void Path(int mk,int x,int y){
if (x == 1 && y == 1) return ;
else if (g[mk][x][y] == 0){
Path(mk,x - 1,y);
printf("D");
}
else{
Path(mk,x,y - 1);
printf("R");
}
}
int main(){
int i,j;
int x;
bool flag = false;
scanf("%d",&n);
for (i = 1; i <= n; ++i){
for (j = 1; j <= n; ++j){
mat[0][i][j] = mat[1][i][j] = 0;
scanf("%d",&x);
if (x == 0){
zx = i;
zy = j;
flag = true;
continue;
}
int tmp = x;
while (tmp % 2 == 0){
mat[0][i][j]++;
tmp /= 2;
}
tmp = x;
while (tmp % 5 == 0){
mat[1][i][j]++;
tmp /= 5;
}
}
}
int ans1 = solve(0);
int ans2 = solve(1);
int ans,mk;
if (ans1 < ans2){
ans = ans1;
mk = 0;
}
else{
ans = ans2;
mk = 1;
}
if (flag&&ans>1){
printf("1\n");
for (i = 2; i <= zx; ++i) printf("D");
for (i = 2; i <= n; ++i) printf("R");
for (i = zx + 1; i <= n; ++i) printf("D");
printf("\n");
}
else{
printf("%d\n",ans);
Path(mk,n,n);
printf("\n");
}
return 0;
}