CF2B The least round way(贪心+动规)

题目

CF2B The least round way

做法

后面\(0\)的个数，\(2\)和\(5\)是\(10\)分解质因数

则把方格中的每个数分解成\(2\)和\(5\)，对\(2\)和\(5\)求两边动规，得出最小值\(ans=min(num_2,num_5)\)

我们贪心地选择最小值所对应的\(2\)或\(5\)，然后从\((n,n)\)按动规路径返回

Code

#include<bits/stdc++.h>
typedef int LL;
const LL maxn=1e3+9;
inline LL Read(){
    LL x(0),f(1); char c=getchar();
    while(c<'0' || c>'9'){
        if(c=='-') f=-1; c=getchar();
    }
    while(c>='0' && c<='9'){
        x=(x<<3)+(x<<1)+c-'0'; c=getchar();
    }return x*f;
}
LL n,m,ax,ay,flag;
LL p[2][maxn][maxn],dp[2][maxn][maxn],a[maxn][maxn];
void Solve(LL x,LL y,LL op){
    if(x==1 && y==1) ;
    else if(x==1){
        Solve(x,y-1,op); printf("R");
    }else if(y==1){
        Solve(x-1,y,op); printf("D");
    }else{
        if(dp[op][x][y-1]==dp[op][x][y]-p[op][x][y]) Solve(x,y-1,op),printf("R");
        else Solve(x-1,y,op),printf("D");
    }
}
int main(){
    n=m=Read();
    for(LL i=1;i<=n;++i)
        for(LL j=1;j<=m;++j){
        	a[i][j]=Read();
        	if(!a[i][j]){
        		ax=i; ay=j;
        		flag=true;
            }
            while(a[i][j]%2==0 && a[i][j]){
                ++p[0][i][j]; a[i][j]/=2;
            }
            while(a[i][j]%5==0 && a[i][j]){
                ++p[1][i][j]; a[i][j]/=5;
            }
        }
    for(LL i=1;i<=n;++i)
        for(LL j=1;j<=m;++j){
        	if(i==1 && j==1){
        		dp[0][i][j]=p[0][i][j];
        		dp[1][i][j]=p[1][i][j];
            }else{
                if(i==1){
                    dp[0][i][j]=dp[0][i][j-1]+p[0][i][j];
                    dp[1][i][j]=dp[1][i][j-1]+p[1][i][j];
                }else if(j==1){
                    dp[0][i][j]=dp[0][i-1][j]+p[0][i][j];
                    dp[1][i][j]=dp[1][i-1][j]+p[1][i][j];
                }else{
                    dp[0][i][j]=std::min(dp[0][i-1][j],dp[0][i][j-1])+p[0][i][j];
                    dp[1][i][j]=std::min(dp[1][i-1][j],dp[1][i][j-1])+p[1][i][j];
                }
            }
        }
    LL ans(std::min(dp[0][n][m],dp[1][n][m]));
    if(ans>1 && flag){
        puts("1");
        for(LL i=1;i<ax;++i) printf("D");
        for(LL i=1;i<ay;++i) printf("R");
        for(LL i=ax;i<n;++i) printf("D");
        for(LL i=ay;i<m;++i) printf("R");
        return 0;
    }
    printf("%d\n",ans);
    if(ans==dp[0][n][m]){
        Solve(n,m,0);
    }else{
        Solve(n,m,1);
    }
    return 0;
}