BUPT Summer Journey #test9 A

 

A. diffsum 2014新生暑假个人排位赛09

时间限制 1000 ms    内存限制 65536 KB    

题目描述

You are givin an array of integers, and you are to figure out the sum of differences between each pair of integers belonging to the array. SEE THE HINT FOR MORE INFORMATION.

输入格式

There are multiple test cases. The first line contains an integer n(n<=1e5), size of the array. The second line contains n integers, ai(|ai|<=100000), the array.

输出格式

For each case, output one line, the answer.

输入样例

41 1 2 2

输出样例

4hintfor the test case the answer is abs(1-1)+abs(1-2)+abs(1-2)+abs(1-2)+abs(1-2)+abs(2-2)=4.

思路：先排序使其具有大小关系。然后进行一轮O(n)的递推即可。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#define maxn 200005//#define LOCALusing namespace std;int n,a[maxn];long long ans;int main(){    #ifdef LOCAL    freopen("input.txt","r",stdin);    #endif // LOCAL    while(scanf("%d",&n)==1)    {        long long sum=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);sum+=a[i];        }        sort(a+1,a+1+n);        sum=sum-(long long)n*(long long)a[1];        ans=sum;        //long long ans=sum;        //printf("%I64d\n",ans);        int j=1;        for(int i=n-1;i>=2;i--)        {            sum=sum+(long long)i*(long long)(a[j]-a[j+1]);            ans+=sum;            j++;        }        printf("%lld\n",ans);    }    return 0;}