POJ 2140 Herd Sums （重要的DP）

最新推荐文章于 2019-11-22 22:57:41 发布

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这篇博客介绍了如何解决POJ 2140题目，重点讨论了动态规划(DP)的应用，解析输入输出格式，并提供样例输入输出以帮助理解解题思路。

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简单题。

题目大意是说求n个牛相互之间距离和，如hint所描述的一样。

首先将这些牛的位置排序。

Description

The cows in farmer John's herd are numbered and branded with consecutive integers from 1 to N (1 <= N <= 10,000,000). When the cows come to the barn for milking, they always come in sequential order from 1 to N.

Farmer John, who majored in mathematics in college and loves numbers, often looks for patterns. He has noticed that when he has exactly 15 cows in his herd, there are precisely four ways that the numbers on any set of one or more consecutive cows can add up to 15 (the same as the total number of cows). They are: 15, 7+8, 4+5+6, and 1+2+3+4+5.

When the number of cows in the herd is 10, the number of ways he can sum consecutive cows and get 10 drops to 2: namely 1+2+3+4 and 10.

Write a program that will compute the number of ways farmer John can sum the numbers on consecutive cows to equal N. Do not use precomputation to solve this problem.

Input

* Line 1: A single integer: N

Output

* Line 1: A single integer that is the number of ways consecutive cow brands can sum to N.

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
__int64 dp[10500];
__int64 a[10500];

int main()
{
	int i,j,n;
	__int64 ans;
	while(scanf("%d",&n)!=EOF){
		for(i=1;i<=n;i++) scanf("%I64d",&a[i]);
		sort(a+1,a+1+n);
		dp[0]=0;
		ans=0;
		for(i=2;i<=n;i++){
			dp[i]=dp[i-1]+(i-1)*(a[i]-a[i-1]);
			ans=dp[i-1]+dp[]
		}
		printf("%I64d\n",ans*2);
	}
	return 0;
}