E - Continued Fractions
A continued fraction of height n is a fraction of form

Input
The first line contains two space-separated integers p, q (1 ≤ q ≤ p ≤ 1018
) — the numerator and the denominator of the first fraction.
The second line contains integer n (1 ≤ n ≤ 90) — the height of the second fraction. The third line contains n space-separated integers a1
, a2
, ..., an
(1 ≤ ai
 ≤ 1018
) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output
Print "YES" if these fractions are equal and "NO" otherwise.

Example
Input
9 422 4

Output
YES

Input
9 432 3 1

Output
YES

Input
9 431 2 4

Output
NO

Note
In the first sample

Test: #44, time: 0 ms., memory: 1844 KB, exit code: 0, checker exit code: 0, verdict: OK
Input
985625905209512860 565433601688714177
10
6423 24947 27507 13031 16414 29169 901 32592 18763 1656
Output
NO
Answer
NO
Checker Log
ok "NO"
题解

#include<stdio.h>
int main()
{
    long long n,m,k,t,temp,i,x,flag;
    scanf("%lld%lld",&n,&m);
    {
        flag=0;
        scanf("%lld",&k);
        for(i=0;i<k;i++)
        {
            scanf("%lld",&t);
            x=n;
            if(n/m<t)
        {
            printf("NO\n");
            return 0;
        }
            n=n-t*m;
            
                if(n==0)
                {
                    break;
                }
            
            temp=m;
            m=n;
            n=temp;
            
        }
        
            if(n==0&&i==k-1)
            {
                printf("YES\n");
            }
            else
            {
                printf("NO\n");
            }
    }
}