标签：alt return memset big tin spec size second integer
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
9 4
2
2 4
YES
9 4
3
2 3 1
YES
9 4
3
1 2 4
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"); } } }
还有C++数学解法:
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‘; } } }
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
