输出无限循环小数的循环节

#include <stdio.h>

void foo( unsigned m, unsigned n )
{
    printf( "%u", m/n );
    m %= n;
    if( m != 0 )
    {
        putchar( '.' );
        unsigned mark[100] = { 0 };
        for( unsigned i=1; m!=0; ++i )
        {
            if( mark[m] != 0 )
            {
                printf( "\nfrom %u to %u", mark[m], i-1 );
                break;
            }
            else
            {
                mark[m] = i;

                putchar( '0' + m*10/n );
                m = m*10%n;
            }
        }
    }
    putchar( '\n' );
    return;
}

int main( void )
{
    foo( 100, 2 );
    foo( 100, 3 );
    foo( 136, 90 );
    foo( 77, 78 );
}