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2017湘潭赛XTU1264Partial Sum

2017-07-13 11:27 253 查看

Partial Sum
Accepted : 155		Submit : 567
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

Sample Output

Source

XTU OnlineJudge

http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1264

想求出前缀和选择连个端点然后累加相当于两个前缀和相减

问题想当与选多对前缀和相减值再相加和最大

先将前缀和排序然后最大减最小第二大减第二小...................

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
ll a[150000];
ll sum[150000];
int main()
{
int n,m;
ll C;
while(~scanf("%d%d%I64d",&n,&m,&C))
{
for(int i=1;i<=n;i++)scanf("%I64d",&a[i]);
for(int i=1;i<=n;i++)
{
if(i==1)sum[i]=a[i];
sum[i]=sum[i-1]+a[i];
}
sort(sum,sum+1+n);
ll cnt=0;
ll ans=0;
for(int i=0;i<m;i++)
{
cnt+=(sum[n-i]-sum[i]-C);
ans=max(ans,cnt);
}
printf("%I64d\n",ans);
}
return 0;
}

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标签： XTU

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