# Luogu P3799 demon dream sticks and p3799 sticks

Source: Internet
Author: User

Luogu P3799 demon dream sticks and p3799 sticks
Background

In the above question, the demon dream chopped a wooden stick, and now she wants to combine the wooden sticks.

Description

There are n wooden sticks. Now, select 4 from them and want to form a triangle. How can I select them?

Input/Output Format

Input Format:

The first line is an integer n.

N integers in the second line, a1, a2 ,...... An (0 <ai <= 5000) represents the length of each wooden rod.

Output Format:

One integer in a row, modulo the 1e9 + 7

Input and Output sample input sample #1: Copy
`4 1 1 2 2`
Output example #1: Copy
`1`
Description

For 30% of Data N <= 5000

For 100% of Data N <= 100000

By-szc

Sure enough, mathematics is still a short board (in fact, I am very spam)

This question should be a bare combination of numbers.

Enumerate short edges and query long edges

Combination quantity pre-processing

`// luogu-judger-enable-o2#include<iostream>#include<cstdio>#define LL long long using namespace std;const LL MAXN=2*1e6+10;const LL mod=1e9+7;LL N;LL a[MAXN],h[MAXN];LL C1(LL a){return a;}LL C2(LL a){return ( a*(a-1) )/2; }int main(){    #ifdef WIN32    freopen("a.in","r",stdin);    #else     #endif    LL ans=0;    scanf("%lld",&N);    for(LL i=1;i<=N;i++) scanf("%lld",&a[i]),h[ a[i] ] ++;    for(LL i=1;i<=5000;i++)        for(LL j=i;j<=5000&&i+j<=5000;j++)        {            if(i==j)             {                if(h[i]>=2&&h[i*2]>=2) ans+=C2(h[i])*C2(h[i*2])%mod;             }            else                 {                 if(h[i]>=1&&h[j]>=1&&h[i+j]>=2)                      ans+=C1(h[i])*C1(h[j])*C2(h[i+j])%mod;            }         }                printf("%lld",ans%mod);    return 0;}`

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