C. Increase Subarray Sums

题意：

f(k)表示对一个数组中不同的k个元素做加x操作，最后的最大字段和。
求f (j) ，1<=j<=n。

思路：

因为n只有5000，那么可以暴力求出长度为len时的最长的字段和，

然后二重循环根据最长字段计算加上能加的值然后输出答案即可。

代码：


#include<cstdio>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
#include <math.h>
#include<vector>
#include<queue>
#include<map>
#define sc_int(x) scanf("%d", &x)
#define sc_ll(x) scanf("%lld", &x)
#define pr_ll(x) printf("%lld", x)
#define pr_ll_n(x) printf("%lld\n", x)
#define pr_int_n(x) printf("%d\n", x)
#define ll long long
using namespace std;const int N=1000000+100;
int n ,m,h;
ll  s[N];
ll sum[N];
ll q[N],ha[N];int main()
{int t;cin>>t;while(t--){cin>>n>>m;for(int i =1;i<=n;i++){cin>>s[i];sum[i]=sum[i-1]+s[i];}for(int i =1;i<=n;i++)ha[i]=-1e18;for(int len =1;len<=n;len++){for(int i=1;i+len-1<=n;i++)ha[len]=max(ha[len],sum[i+len-1]-sum[i-1]);}for(int i =0;i<=n;i++){ll res=0;for(int j =1;j<=n;j++){res=max(res,ha[j]+m*min(i,j));}cout<<res<<" \n"[i==n];}}return 0;
}

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