2017ccpc全国邀请赛(湖南湘潭) E. Partial Sum

E. Partial Sum
Bobo has a integer sequence a 1 ,a 2 ,...,a n of length n. Each time, he selects two ends 0 ≤ l < r ≤ n and
add |
∑ r
j=l+1 a j | − 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 a 1 ,a 2 ,...,a n .
• 2 ≤ n ≤ 10 5
• 1 ≤ 2m ≤ n + 1
• |a i |,C ≤ 10 4
• The sum of n does not exceed 10 6 .
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

官方题解：

前缀和排下序，最大前缀和-最小前缀和就是最大区间和，一直枚举就行了

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define max(x,y) (x)>(y)?(x):(y)
#define min(x,y) (x)>(y)?(y):(x)
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1<<28;
const int maxn=100010;
const int maxm=110;
const int mod=1e9+7;
const double pi=acos(-1.0);
int n,m,c;
int num[maxn],sum[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d%d%d",&n,&m,&c))
    {
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            sum[i]=sum[i-1]+num[i];
        }
        sort(sum,sum+n+1);
        ll ans=0;
        int l=0,r=n;
        for(int i=0;i<m&&l<r;i++)
        {
            if(sum[r]-sum[l]-c>=0)
            {
                ans+=sum[r]-sum[l]-c;
                r--,l++;
            }
        }
        printf("%I64d\n",ans);
    }
    return 0;
}