Moo Volume



H - Moo Volume

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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.

这个题的大意：有N头牛，每只牛都会与另外N-1只牛说话，说话的音量与两只牛之间的距离相等，求这N只牛说话的总音量。

不知道为啥，所有的变量要用long long，否则会WA。

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
long long cow[10010];
int main()
{
    long long n,i;
    scanf("%lld",&n);
    for(i=0;i<n;i++)
        scanf("%lld",&cow[i]);
    sort(cow,cow+n);
    long long sum=0;
    for(i=1;i<n;i++)
        sum+=(cow[i]-cow[i-1])*i*(n-i)*2;        //这是个推导公式，至于如何推导的我也布吉岛为啥
    printf("%lld\n",sum);
    return 0;
}