xtu 1264 Partial Sum 2017湘潭邀请赛E

Partial Sum

Bobo has a integer sequence a1,a2,…,an of length n. Each time, he selects two ends 0≤l<r≤n and add |∑rj=l+1aj|−C into a counter which is zero initially. He repeats the selection for at most m times.

If each end can be selected at most once (either as left or right), find out the maximum sum Bobo may have.

Input

The input contains zero or more test cases and is terminated by end-of-file. For each test case:

The first line contains three integers n, m, C. The second line contains n integers a1,a2,…,an.

• 2≤n≤105
• 1≤2m≤n+1
• |ai|,C≤104
• The sum of n does not exceed 106.

Output

For each test cases, output an integer which denotes the maximum.

Sample Input

4 1 1-1 2 2 -14 2 1-1 2 2 -14 2 2-1 2 2 -14 2 10-1 2 2 -1

Sample Output

3420

#include <stdio.h>#include <algorithm>#include <vector>#include <string.h>#include <queue>#include <cmath>using namespace std; #define ll __int64const int N=100005; int main(){    ll n,m,c,a,sum[N];    while(~scanf("%I64d %I64d %I64d",&n,&m,&c))    {        sum[0]=0;        for(int i=1; i<=n; i++)        {            scanf("%I64d",&a);            sum[i]=sum[i-1]+a;        }        sort(sum,sum+n+1);        int l=n,s=0;        ll ans=0;        for(int i=0; i<m; i++)        {            ll t=abs(sum[n-i]-sum[i])-c;            if(t<=0) break;            ans+=t;        }        printf("%I64d\n",ans);    }    return 0;}