題目
CF2B The least round way
做法
后面\(0\)的個數，\(2\)和\(5\)是\(10\)分解質因數
則把方格中的每個數分解成\(2\)和\(5\)，對\(2\)和\(5\)求兩邊動規，得出最小值\(ans=min(num_2,num_5)\)
我們貪心地選擇最小值所對應的\(2\)或\(5\)，然后從\((n,n)\)按動規路徑返回
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;
}
)
