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

最新推荐文章于 2017-07-18 08:34:59 发布

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最新推荐文章于 2017-07-18 08:34:59 发布

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分类专栏：前缀和文章标签：前缀和

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本篇介绍了一道算法题的解决方案，通过前缀和排序的方法，在给定的整数序列中找到最大可能的累加值。输入包括序列长度、操作次数限制及常数C，输出为最大可能的累加值。

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题目链接：传送门

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

解题思路：前缀和排序

#include <cstdio>  
#include <cstring>  
#include <cmath>  
#include <iostream>  
#include <queue>
#include <set>
#include <string>
#include <stack>
#include <algorithm>
#include <map>
using namespace std;  
typedef  __int64 ll;
const int N = 100100;
const int M = 20;
const int mod = 1e9+7;
const int INF = 0x3fffffff;

ll sum[N];

int main()
{
	int n,m; ll c,a;
	while( ~scanf("%d%d%I64d",&n,&m,&c) ){
		memset( sum , 0 , sizeof(sum) );
		sum[0] = 0;
		for( int i = 1 ; i <= n ; ++i ){
			scanf("%I64d",&a);
			sum[i] += sum[i-1]+a;
		}
		sort( sum , sum+n+1 );
		ll ans = 0;
		for( int i = 0 ; i < m ; ++i ){
			if( (sum[n-i]-sum[i]) <= c ) break;
			ans += (sum[n-i]-sum[i])-c;
		}
		printf("%I64d\n",ans);
	}
	return 0;
}