2017年“嘉杰信息杯” 中国大学生程序设计竞赛全国邀请赛 Partial Sum

Partial Sum
Accepted : 124		Submit : 450
Time Limit : 3000 MS		Memory Limit : 65536 KB

 

Partial Sum

Bobo has a integer sequence a₁,a₂,…,a_n of length n . Each time, he selects two ends 0≤l<r≤n and add |∑rj=l+1a_j|−C into a counter which is zero initially. He repeats the selection for at most mtimes.

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
• |a_i|,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

题意：选取一段区间求和取绝对值，加在初始化为0的数值上，选了的区间不能再选，问最大的和是多少

解法：前缀和排序，最大和最小相减加起来就好了

#include<bits/stdc++.h>
#define  INF 1000000000
#define ll long long
using namespace std;
ll x[123456];
ll sum;
int main()
{
     std::ios::sync_with_stdio(false);
    ll n,m,a,b;
    while(cin>>n>>m>>a)
    {
        sum=0;
        ll ans[123456];
        ans[0]=0;
        for(int i=1;i<=n;i++)
        {
            ll num;
            cin>>num;
            ans[i]=ans[i-1]+num;
        }
        ll cnt=0;
        ll Max=0;
        Max=max(Max,sum);
        sort(ans,ans+1+n);
        while(m--)
        {
            ll Pmax=ans[n--];
            ll Pmin=ans[cnt++];
           // cout<<Pmax<<" "<<Pmin<<endl;
            sum+=abs(Pmax-Pmin)-a;
            if(abs(Pmax-Pmin)-a<0) break;
            Max=max(Max,sum);
        }
        cout<<Max<<endl;
    }
    return 0;
}