17110 Divisible（basic）

17110 Divisible

时间限制:1000MS  内存限制:65535K

题型: 编程题   语言: 无限制

Description

Given n + m integers, I₁,I₂,...,I_n,T₁,T₂,...,T_m, we want to know whether (I1*I2*...*In)%(T1*T2*...*Tm)= =0.

输入格式

The first line gives two integers n and m. 1<=n,m<=100

The second line gives n integers I₁ I₂ ... I_n.

The third line gives m integers T₁ T₂ ... T_n.

1<=I_i, T_i<=2³¹

输出格式

Satisfy (I1*I2*...*In)%(T1*T2*...*Tm)= =0, output "yes", otherwise output "no"

输入样例

2 3
24 14
2 7 3

输出样例

yes

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
ll I[110],T[110];

ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}

int main()
{
    int n, m;
    cin>>n>>m;
    for(int i = 0; i < n; ++i) {
        cin>>I[i];
    }
    for(int i = 0; i < m; ++i) {
        cin>>T[i];
    }

    int i;
    for(i = 0;i < m ; i++){
        for(int j = 0; j < n && T[i] != 1; j++){
            if(I[i]==1) continue;
            ll g=gcd(I[j],T[i]);
            I[j] /= g;
            T[i] /= g;
        }
        if(T[i] != 1) break;//发现一个T[i]不能被I[0]*I[1]*…*I[n]整除,则跳出循环输出no
    }
    if(i < m) cout<<"no";
    else cout<<"yes";
    return 0;
}