描述 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.
约翰的邻居鲍勃控告约翰家的牛们太会叫．
约翰的 N(1≤N≤10000) 只牛在一维的草场上的不同地点吃着草．她们都是些爱说闲话的奶牛，每一只同时与其他 N-1 只牛聊着天．一个对话的进行，需要两只牛都按照和她们间距离等大的音量吼叫，因此草场上存在着 N（N-1）/2 个声音． 请计算这些音量的和．
输入格式 InputFormat
Line 1: N * Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).
第 1 行输入 N，接下来输入 N 个整数，表示一只牛所在的位置．
输出格式 OutputFormat
Line 1: A single integer, the total volume of all the MOOs.
一个整数，表示总音量．
样例输入 SampleInput
5
1
5
3
2
4
样例输出 SampleOutput
40
来源 Source
Silver
代码 Code
递增排序后 \[Ans=\sum_{i=1}^{n-1}i*(n-i)*(a[i+1]-a[i]).\] 注意要用 long long.
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
long long a[10005];
int i,j,t,n,m,l,r,k,z,y,x;
inline int read()
{
int x=0;char ch=getchar();
while (ch<'0' || ch>'9') ch=getchar();
while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
return x;
}
int main()
{
n=read();
for (i=1;i<=n;i++) a[i]=read();
sort(a+1,a+n+1);
long long ans=0;
for (i=1;i<n;i++) ans+=(long long)i*(n-i)*(a[i+1]-a[i]);
ans*=2;
printf("%lld\n",ans);
return 0;
}