题解 | #合唱团#

添加逗号

https://www.nowcoder.com/practice/f51c317e745649c0900996fd3f683aed

思路

线性dp

过程

alt alt

代码

#include <iostream>

using namespace std;

typedef long long LL;

const int N = 60;
const int INF = 0x3f3f3f3f3f3f3f3f;

int a[N], n, k, d;
LL f[N][N], g[N][N];

int main()
{
    cin >> n;
    for(int i = 1;i <= n;i ++) cin >> a[i];
    cin >> k >> d;
    for(int i = 1;i <= n;i ++)
    {
        for(int j = 1;j <= i;j ++)
        {
            if(j == 1) f[i][1] = g[i][1] = a[i];
            else f[i][j] = -INF, g[i][j] = INF;
        }
    }
    for(int i = 2;i <= n;i ++)
    {
        for(int j = 2;j <= i;j ++)
        {
            int l = max(i - d, j - 1), r = i - 1;
            for(int p = l;p <= r;p ++)
            {
                f[i][j] = max(f[i][j], max(f[p][j - 1] * a[i], g[p][j - 1] * a[i]));
                g[i][j] = min(g[i][j], min(f[p][j - 1] * a[i], g[p][j - 1] * a[i]));
            }
        }
    }
    LL ans = -INF;
    for(int i = k;i <= n;i ++) ans = max(ans, f[i][k]);
    cout << ans << endl;
    return 0;
}
全部评论

相关推荐

点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客企业服务