合唱团

题目链接： https://www.nowcoder.com/practice/661c49118ca241909add3a11c96408c8?tpId=85&tqId=29830&tPage=1&rp=1&ru=/ta/2017test&qru=/ta/2017test/question-ranking

解题思路： DP。dp[i][j][0]表示在前i个人里选满足条件的j个人且最后一个人是i的最大值，dp[i][j][1]为对应的最小值。

dp[i][j][0] = max(dp[i-k][j-1][l] * a[i]) 其中k = 1,2，..., D, 和 l = 0,1。dp[i][j][1]有类似的转移方程。

#include <bits/stdc++.h>
using namespace std;

const int imax_n = 55;

int a[imax_n];
long long dp[imax_n][15][2];
int n;
int K;
int D;

int main()
{
    while (scanf("%d", &n)!=-1)
    {
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &a[i]);
        }
        scanf("%d%d", &K, &D);
        dp[0][0][0] = dp[0][0][1] = 1LL;
        for (int i = 1; i <= n; ++i)
        {
            dp[i][0][0] = 1LL;
            dp[i][0][1] = 1LL;
            for (int j = 1; j <= K && j <= i; ++j)
            {
                for (int k = 1; k <= D && i >= k; ++k)
                {
                    for (int l = 0; l <= 1; ++l)
                    {
                        dp[i][j][0] = max(dp[i][j][0], dp[i-k][j-1][l] * a[i]);
                        dp[i][j][1] = min(dp[i][j][1], dp[i-k][j-1][l] * a[i]);
                    }
                }
            }
        }
        long long ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            ans = max(ans, dp[i][K][0]);
        }
        printf("%lld
", ans);
    }
    return 0;
}