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2513: Xiao Yong's academic score (common divisor problem)

2022-07-27 03:08:00 【51CTO】

2513: Xiao Yong's academic score

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 1140

Solved: 794

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Description

Xiao Hui's younger brother Xiao Yong just learned grades , Xiaohui wants to check whether her brother's appointment is correct . Please program to help her .

Input m/n Fractions of form , Output the reduced score （ Only integers that can be divided are output ）

Input

Two Numbers m and n（1<=n,m<=1000）

Output

The score after reduction

Sample Input

Sample Output

HINT

Tips ： Note in the input data '/' The matching of , If the output result is an integer, the denominator is not output

AC Code ：

       
       #include <stdio.h>
       
       #include <stdlib.h>
       
       int 
       
       gcd (
       
       int 
       
       a,
       
       int 
       
       b)
       
{
       
       while(
       
       a
       
       !=
       
       b)
       
    {
       
       if(
       
       a
       
       >
       
       b)
       
       a
       
       =
       
       a
       
       -
       
       b;
       
       if(
       
       b
       
       >
       
       a)
       
       b
       
       =
       
       b
       
       -
       
       a;
       
    }
       
       return 
       
       a;
       
}
       
       int 
       
       main()
       
{
       
       int 
       
       m,
       
       n;
       
       scanf(
       
       "%d/%d",
       
       &
       
       m,
       
       &
       
       n);
       
       int 
       
       q
       
       =
       
       gcd(
       
       m,
       
       n);
       
       int 
       
       x
       
       =
       
       m
       
       /
       
       q,
       
       y
       
       =
       
       n
       
       /
       
       q;
       
       if(
       
       x
       
       %
       
       y
       
       ==
       
       0)
       
       printf(
       
       "%d",
       
       x
       
       /
       
       y);
       
       else
       
       printf(
       
       "%d/%d\n",
       
       x,
       
       y);
       
       return 
       
       0;
       
}
      
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Sum up the least common multiple and the greatest common divisor

       
       /* Minimum common multiple LCM(Lowest Common multiple)
       
        greatest common divisor GCD(greastest common divisor)
       
       */
       
       int 
       
       gcd (
       
       int 
       
       a,
       
       int 
       
       b)
       
       // More subtraction （ in ）
       
{
       
       while(
       
       a
       
       !=
       
       b)
       
    {
       
       if(
       
       a
       
       >
       
       b)
       
       a
       
       =
       
       a
       
       -
       
       b;
       
       if(
       
       b
       
       >
       
       a)
       
       b
       
       =
       
       b
       
       -
       
       a;
       
    }
       
       return 
       
       a;
       
}
       
       int 
       
       gcd(
       
       int 
       
       x,
       
       int 
       
       y)  
       
       // division （ Outside ）
       
{  
       
       int 
       
       r;  
       
       while(
       
       y
       
       >
       
       0)  
       
    {  
       
       r
       
       =
       
       x
       
       %
       
       y;  
       
       x
       
       =
       
       y;  
       
       y
       
       =
       
       r;  
       
    }  
       
       return 
       
       x;  
       
}  
       
       int 
       
       lcm(
       
       int 
       
       x,
       
       int 
       
       y)  
       
{  
       
       return 
       
       x
       
       *
       
       y
       
       /
       
       gcd(
       
       x,
       
       y);  
       
}
      
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