Continued Fractions CodeForces – 305B (java+高精 / 数学)

发表于:魔客站 2019-4-15 分类:建站教程 阅读: 18

标签:alt   return   memset   big   tin   spec   size   second   integer   

A continued fraction of height n is a fraction of form 技术分享图片. You are given two rational numbers, one is represented as 技术分享图片 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 技术分享图片.

In the second sample 技术分享图片.

In the third sample 技术分享图片.

 

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:

技术分享图片技术分享图片

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");        }            }}

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还有C++数学解法:

我们看一个简单的等式:

技术分享图片

把它上下翻转一下呢?

技术分享图片

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:

技术分享图片技术分享图片

#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;        }    }}

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Continued Fractions CodeForces - 305B (java+高精 / 数学)

标签:alt   return   memset   big   tin   spec   size   second   integer   

原文地址:https://www.cnblogs.com/qieqiemin/p/10339938.html