Codeforces Round #184 (Div. 2)-B. Continued Fractions

B. Continued Fractions

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

A continued fraction of height n is a fraction of form 

.
You are given two rational numbers, one is represented as 

 and
the other one is represented as a finite fraction of height n. Check if they are equal.

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 integersa1, 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 %I64dspecifier.

Output

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

Sample test(s)

input

9 4
2
2 4

output

YES

input

9 4
3
2 3 1

output

YES

input

9 4
3
1 2 4

output

NO

Note

In the first sample 

.

In the second sample 

.

In the third sample 

.
这道题，真心没什么好说的，直接正向模拟，不过要注意q==0时要跳出循环，刚做这道题时，就想从后往前推，结果不论怎样就是WA和ET，所以左后还是没做出来，这还是看了别人的思路才敲出来的，废话不多说，直接看代码吧：
#include<iostream>

#include<algorithm>

using namespace std;

typedef long long s;//typedef用来为复杂的声明定义简单的别名

int main()

{
s q,p,n,m,k,a;
cin>>p>>q;
cin>>n;
k=0;
while(n)
{
cin>>m;
if(q<=0||m>(p/q))//这里要先判断q是否为0，为0要跳出循环
{
k=1;
break;
}
p=p-m*q;//关键，模拟
swap(p,q);
n--;
}
if(k!=1)
{
if(q==0)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
else
cout<<"NO"<<endl;
return 0;

}