boj problem 1340 简单题 费时很久 long long型 输入输出时 "%lld“ 是LL而不是11. 先快排 然后计算就可以

地址：http://acm.scs.bupt.cn/onlinejudge/showproblem.php?problem_id=1340

The Noisy PartySubmit: 209 Accepted:50Time Limit: 1000MS Memory Limit: 65536K
Description
Nitaa is holding a party at his house. But unfortunately Nitaa has received a noise complaint from his neighbor, Arsenal4, stating that his friends in his room are making too much noise.

Nitaa's N friends (1 <= N <= 10,000) all sit at various locations on a long one-dimensional pasture. The friends are very chatty people. Every pair of friends simultaneously carries on a conversation (so every friend is simultaneously talking with all of the N-1 other friends). When friend i talks with friend j, the volume of this talk must be equal to the distance between i and j, in order for j to be able to hear the conversation at all. Please help Nitaa compute the total volume of sound being generated by all N*(N-1) simultaneous conversation. That is the total volume of conversation between all pairs of friends.

Input
* Line 1: N

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

Output
Only one line that shows the total volume of the noise.

Sample Input

5
1
5
3
2
4

Sample Output

40

Source

#include<iostream>
#include<math.h>
using namespace std;
int quicksort(long long *a,int low,int high)
{
long long temp;
int mid;
mid=(low+high)/2;
temp=a[mid];
a[mid]=a[low];
a[low]=temp;
while(low<high)
{
while(low<high && a[high]>temp)
high--;
if(low<high)
a[low++]=a[high];
while(low<high && a[low]<temp)
low++;
if(low<high)
a[high--]=a[low];

}
a[low]=temp;
return low;
}
void qsort(long long *a,int low,int high)
{
int temp;
if(low<high)
{
temp=quicksort(a,low,high);
qsort(a,low,temp-1);
qsort(a,temp+1,high);
}
}

int main()
{
long long a[10001],b[10001];
int N, i;
long long sum=0;
scanf( "%d", &N );
for( i=1; i<=N; i++)
cin>>a[i];
//scanf("%lld",a+i);

qsort(a,1,N);
b[1]=0; b[2]=a[2]-a[1];
sum=b[1]+b[2];
for(i=3;i<=N;i++)
{
b[i]=(a[i]-a[i-1])*(i-1)+b[i-1];
sum +=b [i];

}
sum=sum*2;
cout<<sum<<endl;
// printf( "%lld/n", sum*2 );

}