2017-5-14 湘潭市赛 Partial Sum 给n个数，每次操作选择一个L,一个R，表示区间左右端点，该操作产生的贡献为[L+1,R]的和的绝对值-C。 0<=L<R<=n; 如果选过L,R这两个位置，那么以后选择的L,R都不可以再选择这两个位置。最多操作m次，求可以获得的 最大贡献和。

Partial Sum
Accepted : 30           Submit : 119
Time Limit : 3000 MS           Memory Limit : 65536 KB

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

Sample Output

3
4
2
0


Source
XTU OnlineJudge 

/**
题目：Partial Sum
链接：http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1264
题意：给n个数，每次操作选择一个L,一个R，表示区间左右端点，该操作产生的贡献为[L+1,R]的和的绝对值-C。
0<=L<R<=n; 如果选过L,R这两个位置，那么以后选择的L,R都不可以再选择这两个位置。最多操作m次，求可以获得的
最大贡献和。
思路：脑洞。原公式可以转化为|sum[r]-sum[l]|-c，sum[i]表示前i项的前缀和。
由于绝对值的影响，所以对前缀和排序，然后l从左边开始递增枚举，r从右边开始递减枚举，每次组成一对(sum[l],sum[r])作为贡献计算。


*/

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;
const int maxn = 1e5+100;
int a[maxn], sum[maxn];
int main()
{
    int n, m, c;
    while(scanf("%d%d%d",&n,&m,&c)!=EOF)
    {
        //cout<<"yes"<<endl;
        for(int i = 1; i <= n; i++) scanf("%d",&a[i]);
        sum[0] = 0;
        for(int i = 1; i <= n; i++){
            sum[i] = a[i]+sum[i-1];
        }
        sort(sum,sum+1+n);
        int l = 0, r = n;
        LL ans = 0;
        while(m--){
            int value = abs(sum[r]-sum[l]);
            if(value<=c) break;
            ans += value-c;
            l++, r--;
        }
        printf("%lld
",ans);
    }
    return 0;
}