（Problem 33）Digit canceling fractions

The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

题目大意：

分数 49/98 是一个奇怪的分数：当一个菜鸟数学家试图对其进行简化时，他可能会错误地可以认为通过将分子和分母上的9同时去除得到 49/98 = 4/8。但他得到的结果却是正确的。

我们将30/50 = 3/5这样的分数作为普通个例。

一共有四个这样的非普通分数，其值小于1，并且包括分子和分母都包括2位数。 如果将这四个分数的乘积约分到最简式，分母是多少？

//（Problem 33）Digit canceling fractions
// Completed on Thu, 25 Jul 2013, 17:47
// Language: C
//
// 版权所有（C）acutus   (mail: acutus@126.com)
// 博客地址：http://www.cnblogs.com/acutus/

#include<stdio.h>
void swap(int *a, int *b)
{
int t;
t=*a;
*a=*b;
*b=t;
}

int gcd(int a, int b)
{
int r;
if (a < b)
swap(&a,&b);
if (!b)
return a;
while ((r = a % b) != 0) {
a = b;
b = r;
}
return b;
}

void find()
{
int i;
int M,N;
M=N=1;
for(i=12; i<50; i++)
{
for(int j=i+1; j<100; j++)
{
int t=gcd(i,j);
if(t==1 || i/t>10 || j/t>10 || i%10!=j/10)
continue;
else
{
int a=i/10,b=j%10;
if(a/gcd(a,b)==i/t && b/gcd(a,b)==j/t)
{
M*=i/t;
N*=j/t;
}
}
}
}
printf("%d\n",N/gcd(M,N));
}

int main()
{
find();
return 0;
}

Answer:

100