POJ2231总结

Moo Volume

题目来源

Description

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise.

FJ’s N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.

Input

* Line 1: N

* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).

Output

There are five cows at locations 1, 5, 3, 2, and 4.

Sample Input

5
1
5
3
2
4

Sample Output

40

Hint

INPUT DETAILS:

There are five cows at locations 1, 5, 3, 2, and 4.

OUTPUT DETAILS:

Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.

我的代码

两只牛之间叫一次需要被计算两次，现在就先只计算一段距离，结果乘以二就是最终答案。

先对所有牛按照坐标升序排序，每次加入一头牛。在第i头牛加入战场时，它造成的噪音是(i-1)*它的坐标-前（i-1）头牛的坐标之和。这个证明画个图就出来了。

#include<stdio.h>
#include<algorithm>

using namespace std;

int locations[1000010];
long long sum;
long long ans;
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=0;i<n;i++){
		scanf("%d",&locations[i]);
	}
	sort(locations,locations+n);
	sum=locations[0];
	ans=0;
	//要开long long
	for (long long int i=1;i<n;i++){
		ans+=i*locations[i]-sum;
		sum+=locations[i];
	}
	printf("%lld\n",ans*2);
	return 0;
}

反思

这个数据量很大，实际上要开longlong才能过。