2017湘潭赛Partial Sum_partial sum acm-优快云博客

本文链接：https://blog.youkuaiyun.com/xky140610205/article/details/74852961

本文解析了一个关于整数序列的问题——PartialSum，旨在通过选择特定区间并计算其和，然后减去常数C来最大化总和。限制条件为每个端点只能使用一次，且操作次数有限。文章提供了一种解决方案，并附带了示例输入输出。

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   Partial Sum 
Accepted : 155 Submit : 567
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
 
     

 Source XTU OnlineJudge
 

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

给一个序列，从中取一段，求和再减去C得到一个值，最多重复m次，求这些值的和（最大），每个端点只能取一次；

一道思维题大神博客 http://blog.youkuaiyun.com/mengxiang000000/article/details/72228834

#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);
    }
}