hdu 5912(迭代+gcd)

最新推荐文章于 2024-08-03 08:00:00 发布

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最新推荐文章于 2024-08-03 08:00:00 发布

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文章标签： java 开发工具

原文链接：http://www.cnblogs.com/liyinggang/p/5931382.html

本文介绍了一种计算复杂分数表达式的算法，并通过迭代求解和约分的方式给出正确的结果。文章提供了一个完整的C++实现案例。

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Fraction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 97 Accepted Submission(s): 64

Problem Description

Mr. Frog recently studied how to add two fractions up, and he came up with an evil idea to trouble you by asking you to calculate the result of the formula below:

As a talent, can you figure out the answer correctly?

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains only one integer n (

Output

For each case, print a line “Case #x: p q”, where x is the case number (starting from 1) and p/q indicates the answer.

You should promise that p/q is irreducible.

Sample Input

Sample Output

Case #1: 1 2
Hint

Here are the details for the first sample:
2/(1+3/1) = 1/2

题意:求图上的公式的结果.直接迭代求然后约分

#include <bits/stdc++.h>
using namespace std;
map<int ,int> value;
int a[10],b[10];
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}
int main()
{
    int tcase,t=1;
    scanf("%d",&tcase);
    while(tcase--){
        int n,res1,res2;
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=n;i++) scanf("%d",&b[i]);
        res1 = a[n],res2 = b[n];
        for(int i=n-1;i>=1;i--){
            int k = res1;
            res1 = a[i]*res1+res2;
            res2 = b[i]*k;
        }
        int d = gcd(res1,res2);
        printf("Case #%d: %d %d\n",t++,res2/d,res1/d);
    }
    return 0;
}
 