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.
求个和，简单题
#include<stack>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=100005;
ll T,n,m,a[maxn],ans,sum;
int main()
{
while (~scanf("%lld",&n))
{
ans=sum=0;
for (ll i=1;i<=n;i++)
{
scanf("%lld",&a[i]);
sum+=a[i];
}
sort(a+1,a+n+1);
for (ll i=1;i<=n;i++)
{
ans+=sum-a[i]*(n-i+1);
sum-=a[i];
}
printf("%lld\n",ans+ans);
}
return 0;
}