C++11(clang++ 3.9) 解法, 执行用时: 139ms, 内存消耗: 484K, 提交时间: 2019-10-05 22:09:12

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const LL MOD = 1e9 + 7;

LL qpow(LL x, LL y){
	LL z = 1;
	while (y > 0){
		if (y & 1) z = z * x % MOD;
		x = x * x % MOD;
		y /= 2;
	}
	return z;
}

int main(){
	int n;
	while (scanf("%d", &n) != EOF){
		LL ans1 = 1, ans2 = 1, P = 1e10 + 19; 
		for (int i = 1; i <= n; i++){
			int x; scanf("%d", &x);
			ans1 = ans1 * (P % MOD - x + MOD) % MOD;
			ans2 = ans2 * (MOD - x) % MOD;
		}
		LL ans = (ans1 - ans2 + MOD) * qpow(P % MOD, MOD - 2) % MOD;
		printf("%lld\n", ans);
	}
	return 0;
}

C++14(g++5.4) 解法, 执行用时: 541ms, 内存消耗: 468K, 提交时间: 2020-05-21 18:08:46

#include<iostream>
using namespace std;
#define int long long
int mod=1e9+7;
int p=1e10+19;
int n;

int qpow(int a,int b){
    int ans=1;
    for(;b;b>>=1){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
    }
    return ans;
}


signed main()
{
    p%=mod;
    while(cin>>n){
        int ans1=1,ans2=1;
        for(int i=1;i<=n;i++){
            int t;
            cin>>t;
            ans1=ans1*(p-t)%mod;
            ans2=ans2*(mod-t)%mod;
        }
        cout<<(ans1-ans2+mod)%mod*qpow(p,mod-2)%mod<<endl;
    }
    return 0;
}