RealPhobia

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 69    Accepted Submission(s): 27

Problem Description Bert is a programmer with a real fear of floating point arithmetic. Bert has quite successfully used rational numbers to write his programs but he does not like it when the denominator grows large. Your task is to help Bert by writing a program that decreases the denominator of a rational number, whilst introducing the smallest error possible. For a rational number A/B, where B > 2 and 0 < A < B, your program needs to identify a rational number C/D such that:

1. 0 < C < D < B, and

2. the error |A/B – C/D| is the minimum over all possible values of C and D, and

3. D is the smallest such positive integer.  

Input The input starts with an integer K (1 <= K <= 1000) that represents the number of cases on a line by itself. Each of the following K lines describes one of the cases and consists of a fraction formatted as two integers, A and B, separated by “/” such that:

1. B is a 32 bit integer strictly greater than 2, and

2. 0 < A < B  

Output For each case, the output consists of a fraction on a line by itself. The fraction should be formatted as two integers separated by “/”.  

Sample Input 3 1/4 2/3 13/21  

Sample Output 1/3 1/2 8/13  

Source
The 2011 South Pacific Programming Contest  

Recommend lcy  

分解成连分数，然后最后一个数减一；

//HDU 4180 连分数 

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int an[100];
int gcd(int a,int b)
{
    int r=0;
    while(b!=0)
    {
        r=a%b;
        a=b;
        b=r;
    }    
    return a;
}    
int fenjie(int a,int b,int an[])//连分数分解 
{
    int n=0;
    int t;
    while(a!=1)
    {
        an[n++]=b/a;
        t=b%a;
        b=a;
        a=t;
    }    
    an[n++]=b;
    return n;
}    
int main()
{
    int n,A,B,C,D;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d/%d",&A,&B);
            int d=gcd(A,B);
            if(d>1)
            {
                printf("%d/%d\n",A/d,B/d);
                continue;
            }  
            int len=fenjie(A,B,an); 
            an[len-1]--;
            C=1;
            D=an[len-1];
            for(int j=len-2;j>=0;j--)
            {
                int t=an[j]*D+C;
                C=D;
                D=t;
            }    
            printf("%d/%d\n",C,D);
            
        }    
        
    }   
    return 0; 
}