The least round way

给你一个n*n的矩阵，求从左上角到右下角的所有路径中，最后乘机末尾0最小的路径，输出末尾0的个数和路径

题解

将每个数提取质因数2和5，然后dp每个点2和5的个数，最后取两个中最小的，如果矩阵中有0， 且5和2的最小值不为0，则随便输出一个经过0的路径

#include <iostream>
#include <cstdio>

using namespace std;

const int N=1005;
int l[N][N][3],dp[N][N][3];
char pre[N][N][2];

int get2(int x)
{
    int res = 0;
    while(x % 2 == 0)
    {
        res++;
        x /= 2;
    }
    return res;
}

int get5(int x)
{
    int res = 0;
    while(x % 5 == 0)
    {
        res ++;
        x /= 5;
    }
    
    return res;
}

void print(int x,int y,int q)
{
    if(x==1 && y==1) return ;
    if(pre[x][y][q] == 'R'){
        print(x,y-1,q);
        printf("R");
    }
    else{
        print(x-1,y,q);
        printf("D");
    }
}

int main()
{
    int n,a,x=-1,y=-1;
    cin >> n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            cin >> a;
            if(a == 0){
                l[i][j][0] = 0;
                l[i][j][1] = 0;
                x = i;
                y = j;
            }
            
            else{
                l[i][j][0] = get2(a);
                l[i][j][1] = get5(a); 
            }
        }
    }

    for(int i = 1;i <=n ;i++){
        dp[1][i][0] = dp[1][i-1][0] + l[1][i][0];
        dp[1][i][1] = dp[1][i-1][1] + l[1][i][1];
        pre[1][i][0] = 'R';
        pre[1][i][1] = 'R';
        
        dp[i][1][0] = dp[i-1][1][0] + l[i][1][0];
        dp[i][1][1] = dp[i-1][1][1] + l[i][1][1];
        pre[i][1][0] = 'D';
        pre[i][1][1] = 'D';
    }

    
    for(int i = 2 ; i <= n ;i++){
        for(int j = 2; j <= n; j++){
            if(dp[i-1][j][0] < dp[i][j-1][0]){
                dp[i][j][0] = dp[i-1][j][0] + l[i][j][0]; 
                pre[i][j][0] = 'D';
            }
            else{
                dp[i][j][0] = dp[i][j-1][0] + l[i][j][0];
                pre[i][j][0] = 'R';
            }
            if(dp[i-1][j][1] < dp[i][j-1][1]){
                dp[i][j][1] = dp[i-1][j][1] + l[i][j][1];
                pre[i][j][1] = 'D';
            }
            else{
                dp[i][j][1] = dp[i][j-1][1] + l[i][j][1];
                pre[i][j][1] = 'R';
            }
        }
    }
    
    if(x==-1){
        if(dp[n][n][0] < dp[n][n][1]){
            cout << dp[n][n][0] << endl;
            print(n,n,0);
        }
        else{
            cout << dp[n][n][1] << endl;
            print(n,n,1);
        }
    }
    else if(x!=-1 && min(dp[n][n][0],dp[n][n][1]) == 0){
        cout << "0" << endl;
        if(dp[n][n][0] == 0) print(n,n,0);
        else print(n,n,1);
    }
    else{ 
        cout << "1" << endl;
        for(int i=1;i<=n;i++){
            if(i==x){
                for(int j=1;j<n;j++){
                    printf("R");
                }
            }
            else{
                printf("D");
            }
        }
    }
    
    return 0;
}