Description
Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p mod m is defined as the smallest non-negative integer k such that p = a*m + k for some integer a.
Input
Input consists of a number of cases. Each case is represented by a line containing three unsigned integers. The first, b, is a decimal number between 2 and 10. The second, p, contains up to 1000 digits between 0 and b-1. The third,
m, contains up to 9 digits between 0 and b-1. The last case is followed by a line containing 0.
m, contains up to 9 digits between 0 and b-1. The last case is followed by a line containing 0.
Output
For each test case, print a line giving p mod m as a base-b integer.
Sample Input
2 1100 101 10 123456789123456789123456789 1000 0
Sample Output
10 789
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void print(int r,int b)
{
if(r>0)
{
print(r/b,b);
printf("%d",r%b);
}
}
int main()
{
char p[10000],M[10];
int n,i,m,r;
while(~scanf("%d",&n))
{
if(!n)
return 0;
scanf("%s%s",p,M);
int len = strlen(p);
m = strtol(M,0,n);
r = 0;
for(i = 0; i<len; i++)
{
r = r*n+p[i]%48;
r%=m;
}
print(r,n);
puts(r?"":"0");
}
return 0;
}