XTUOJ1264：Partial Sum（前缀和）

Partial Sum

Accepted : 4 Submit : 12Time 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 -14 2 1-1 2 2 -14 2 2-1 2 2 -14 2 10-1 2 2 -1

Sample Output

3420

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;}