洛谷 P1043 数字游戏

题目传送门

解题思路:

跟石子合并差不多,区间DP(环形),用f[i][j][s]表示从i到j分成s段所能获得的最大答案,枚举断点k,则f[i][j][s] = min(f[i][j][s],f[i][k][s-1] * 代价),最小值反之.

ps:区间和用前缀和维护.

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n,m,sum[120],f[200][200][10],f2[200][200][10],a[52];

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n; i++) {
        scanf("%d",&a[i]);
        sum[i] = sum[i-1] + a[i];
    }
    memset(f,0x3f3f,sizeof(f));
    for(int i = n + 1;i <= 2 * n; i++)
        sum[i] = sum[i-n] + sum[n];
    for(int i = 1;i <= 2 * n; i++)
        for(int j = i;j <= 2 * n; j++)
            f[i][j][1] = f2[i][j][1] = ((sum[j] - sum[i-1]) % 10 + 10) % 10;
    for(int len = 0;len < n; len++)
        for(int i = 1,j = i + len;j <= 2 * n; i++,j++) 
            for(int s = 2;s <= m; s++)
                for(int k = i + s - 1;k < j; k++) {
                    f[i][j][s] = min(f[i][j][s],f[i][k][s-1] * (((sum[j] - sum[k]) % 10 + 10) % 10));
                    f2[i][j][s] = max(f2[i][j][s],f2[i][k][s-1] * (((sum[j] - sum[k]) % 10 + 10) % 10));
                }
    int maxans = 0;
    int minans = 0x3f3f3f3f;
    for(int i = 1;i <= 2 * n; i++) {
        maxans = max(maxans,f2[i][i+n-1][m]);
        minans = min(minans,f[i][i+n-1][m]);
    }
    printf("%d\n%d",minans,maxans);
    return 0;
}