RealPhobia

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 376    Accepted Submission(s): 151

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

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#include<stdio.h>
#include<string.h>

long long gcd1(long long a,long long b,long long &x,long long &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    long long d = gcd1(b,a%b,x,y);
    long long t = x;
    x = y;
    y = t - a/b*y;
    return d;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long a,b;
        scanf("%lld/%lld",&a,&b);
        long long x = 0,y = 0;
        long long p = gcd1(a,b,x,y);
        //printf("%lld,%lld\n",x,y);
        //printf("---%lld\n",p);
        //printf("==%lld %lld\n",a,b);
        if(p != 1)
        {
            printf("%lld/%lld\n",a/p,b/p);
            continue;
        }
        if(a == 1)
        {
            printf("1/%lld\n",b-1);
            continue;
        }
        long long x1 = 0,y1 = 0;
        if(x > 0)
        {
            x1 = (a + y)%a;
            y1 = (b - x)%b;
        }
        else
        {
            x1 = (a - y)%a;
            y1 = (b + x)%b;
        }
        //printf("%lld %lld %lld %lld\n",x1,y1);
        printf("%lld/%lld\n",x1,y1);
    }
    return 0;
}