How mang NFind a minimal interger K which is merely comprise
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How mang N
Find a minimal interger K which is merely comprised of N and can be divided by M.
For example,11 is the minimal number that and be divided by 11, and it is comprised of two '1's, and 111111 can be divided by 13 which is comprised of six '1's.
Input
On each line of input , there will be two positive integer, N and M. N is a digit number, M is no more than 10000.
Output
On each single line, output the number of N, if no such K, output zero.
Sample Input
1 5
1 11
1 13
Sample Output
0
2
6
Find a minimal interger K which is merely comprised of N and can be divided by M.
For example,11 is the minimal number that and be divided by 11, and it is comprised of two '1's, and 111111 can be divided by 13 which is comprised of six '1's.
Input
On each line of input , there will be two positive integer, N and M. N is a digit number, M is no more than 10000.
Output
On each single line, output the number of N, if no such K, output zero.
Sample Input
1 5
1 11
1 13
Sample Output
0
2
6
哪来的题目,描述真是破绽百出.
#include
int main( )
{
int n,m,t,f[ 10000 ],i,count;
while ( scanf("%d%d",&n,&m) == 2 )
{
count = t = 0;
for ( i = 0; i < m; i++ )
f[ i ] = 0;
while ( !f[ t ] )
{
count++;
f[ t ] = 1;
t = ( t * 10 + n ) % m;
}
if ( !t )
printf("%d\n",count);
else
printf("0\n");
}
return 0;
}
#include
int main( )
{
int n,m,t,f[ 10000 ],i,count;
while ( scanf("%d%d",&n,&m) == 2 )
{
count = t = 0;
for ( i = 0; i < m; i++ )
f[ i ] = 0;
while ( !f[ t ] )
{
count++;
f[ t ] = 1;
t = ( t * 10 + n ) % m;
}
if ( !t )
printf("%d\n",count);
else
printf("0\n");
}
return 0;
}
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