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[],T[];

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

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

     int i;
     for(i = ;i < m ; i++){
         for(int j = ; j < n && T[i] != ; j++){
             if(I[i]==) continue;
             ll g=gcd(I[j],T[i]);
             I[j] /= g;
             T[i] /= g;
         }
         if(T[i] != ) 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 ;
 }