2017四川省赛L题Nice Trick

题目链接：​​省赛PDF​​

题目大意：题目给了你从任意数中选出三个数，然后相乘，并加到总和的公式，现在要你从任意数中选出四个数，相乘，并加到总和，问最后的答案，对1e9+7取模

题目思路：枚举第四个数，对它前面的数套公式就好，除6的时候写一下逆元（大一的学弟不会逆元，死都没过掉，所以这种算法题还是不要让他们写的好？）

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
const int MOD = 1e9+7;

ll n,a[maxn],sum1[maxn],sum2[maxn],sum3[maxn],inv[6];

ll quick(ll a,ll b,ll mod)
{
    ll ans = 1;
    while(b)
    {
        if(b&1)
            ans = (ans*a)%mod;
        a = (a*a)%mod;
        b >>= 1;
    }
    return ans;
}

ll solve(ll x){
    ll s1 = (((sum1[x]*sum1[x])%MOD)*sum1[x])%MOD;
    ll s2 = (((sum2[x]*sum1[x])%MOD)*3)%MOD;
    ll s3 = (2*sum3[x])%MOD;
    ll s4 = s1-s2+s3;
    ll s5 = (s4*quick(6,MOD-2,MOD))%MOD;
    return s5;
}

int main(){
    while(~scanf("%lld",&n)){
        sum1[0] = sum2[0] = sum3[0] = 0;
        for(int i = 1;i <= n;i++){
            scanf("%lld",&a[i]);
            sum1[i] = (sum1[i-1]+a[i])%MOD;
            sum2[i] = (sum2[i-1]+(a[i]*a[i])%MOD)%MOD;
            sum3[i] = (sum3[i-1]+(((a[i]*a[i])%MOD)*a[i])%MOD)%MOD;
        }
        ll sum = 0;
        for(int i = 4;i <= n;i++){
            sum = (sum + (a[i] * solve(i-1))%MOD)%MOD;
        }
        printf("%lld\n",sum%MOD);
    }
    return 0;
}