XTUOJ1264：Partial Sum（前缀和）-优快云博客

本文探讨了一个关于整数序列的算法问题，旨在通过选择特定的子序列并计算其绝对值之和减去常数C的方式，找出最大的可能累加值。文章提供了详细的解决思路与C++代码实现。

Partial Sum
Accepted : 4 Submit : 12
Time Limit : 3000 MS Memory Limit : 65536 KB 

Partial SumBobo 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.

InputThe 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.

OutputFor 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

思路：先统计前缀和，由于和取绝对值，所以可以直接进行排序，又每个点只能选一次，首尾两个指针同时向中间靠即可，遇到<C的直接退出。

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
using namespace std;

const int N = 100030;
typedef long long LL;
LL a[N];
int main()
{
    LL n, m, c, t;
    while(~scanf("%I64d%I64d%I64d",&n,&m,&c))
    {
        a[0] = 0;
        for(LL i=1; i<=n; ++i)
        {
            scanf("%I64d",&t);
            a[i] = a[i-1] + t;
        }
        sort(a, a+1+n);
        LL sum=0, cnt=0;
        for(LL i=0,j=n; i<j&&cnt<m ; ++i,--j)
        {
            LL tmp = a[j]-a[i];
            if(tmp >= c)
                sum += tmp,++cnt;
            else
                break;
        }
        printf("%I64d\n",sum-cnt*c);
    }
    return 0;
}