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 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.

Examples

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 .

 

思路：

可以用java的高精度类BigDecimal直接暴力模拟分式的递归运算过程，精度保留到30以上均可以AC

我的JAVA代码：

import java.math.*;
import java.util.Scanner;
public class Main {
    
    static long  a[] = new long[500];
    public static int n;
    public static BigDecimal f(int index)
    {
        if(index==n)
            return BigDecimal.valueOf(a[index]).divide(BigDecimal.ONE,45,BigDecimal.ROUND_HALF_DOWN);
        else
            return BigDecimal.valueOf(a[index]).add(BigDecimal.ONE.divide(f(index+1),45,BigDecimal.ROUND_HALF_DOWN));
    }
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigDecimal p,q;
        
        p=cin.nextBigDecimal();
        q=cin.nextBigDecimal();
        n=cin.nextInt();
        for(int i=1;i<=n;i++)
        {
            a[i]=cin.nextLong();
        }
        BigDecimal s1=p.divide(q,45,BigDecimal.ROUND_HALF_DOWN);
        BigDecimal s2=f(1);
//        System.out.println(s1);
//        System.out.println(s2);
        if(s1.equals(s2))
        {
            System.out.println("YES");
        }else
        {
            System.out.println("NO");
        }
        
    }

}
View Code

还有C++数学解法:

我们看一个简单的等式：

把它上下翻转一下呢？

这样迭代下去，如果是相等的，那么右边一定是等于0的.

图来自这位大佬的博客：

https://blog.csdn.net/theArcticOcean/article/details/50429314

注意：

中间特判下分母为0的情况和中途出结果的情况

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-16
#define eps2 1e-15
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll p,q;
int n;
ll a[maxn];
int main()
{
    dll(p);dll(q);
    gg(n);
    repd(i,1,n)
    {
        dll(a[i]);
    }
    int flag=0;
    repd(i,1,n)
    {
        if(q==0||(p*1.0000000/q)<a[i])
        {
            flag=1;
            break;
        }else
        {
            p-=q*a[i];
            swap(p,q);
        }
    }
    if(flag==0&&q==0)
    {
        printf("YES\n");
    }else
    {
        printf("NO\n");
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}
View Code