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
* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).
Output
Sample Input
5
1
5
3
2
4
Sample Output
40
Hint
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.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;//要找规律！！！
ll a[11000];
int main()
{
int n;
cin>>n;
for(int i=0;i<n;i++) scanf("%I64d",&a[i]);
sort(a,a+n);
ll sum=0;
for(int i=1;i<n;i++)
{
sum+=(a[i]-a[i-1])*i*(n-i)*2;
}
cout<<sum<<endl;
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
ll a[11000],b[11000];
int main()
{
int n;
cin>>n;
for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
sort(a+1,a+n+1);
for(int i=2;i<=n;i++)
{
b[1]+=abs(a[i]-a[1]);
}
ll sum=b[1];
for(int i=2;i<=n;i++)
{
ll d=a[i]-a[i-1];
b[i]=b[i-1]+(i-1-1)*d-(n-i)*d;
sum+=b[i];
}
cout<<sum<<endl;
return 0;
}