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技术 2022年11月23日
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Moo Volume

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 22104   Accepted: 6692

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.

————————————————————————————————————

题目的意思是给出N点,,算出每个点到其他点的距离和,再求和

找到规律推出公式

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include<vector>
#include <map>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long longLL a[10005];int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
sort(a,a+n);
LL ans=0;
for(int i=1;i<n;i++)
{
ans+=(a[i]-a[i-1])*i*(n-i)*2;
}
printf("%lld\n",ans); } return 0;
}

跑暴力貌似也能过

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <string>
#include <set>
#include<vector>
#include <map>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long longLL a[10005];int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
sort(a,a+n);
LL ans=0;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
ans+=(a[j]-a[i]);
}
printf("%lld\n",ans*2); } return 0;
}

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