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[],f[][][],f2[][][],a[];

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