Big Number

Big Number 

Problem Description

As we know, Big Number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

Output

For each test case, you have to ouput the result of A mod B.

Sample Input

2 3
12 7
152455856554521 3250

Sample Output

2
5
1521

 

#include<stdio.h>
#include<string.h>
int b1,b2,b3,a1[110000],a2[110000],n;
char str1[110000],str2[110000],str3[110000],*p;
int main()
{ 
void fun();
p=str1;
    int i,j,k,x=0,c;
   while( gets(str1))
   { 
   for(i=0;i<11000;i++)
       { if(str1[i]==' ')
        {str1[i]='';
  strcpy(str2,p+i+1);break;}}
     
     
    b1=strlen(str1);
    b2=strlen(str2);
    if(b1<b2){puts(str1); }
      else if(b1==b2&&(strcmp(str1,str2)<0)){puts(str1);}
    else
    {int h=b2-1;
    for(i=b2-1;i>=0;i--)
      a1[b2-i-1]=str1[i]-'0';
    for(i=b2-1;i>=0;i--)
      a2[b2-i-1]=str2[i]-'0';
    for(j=0;j<=b1-b2;j++)
    {
      fun();
       h++;
           for(i=b2;i>0;i--)
           a1[i]=a1[i-1];
           a1[0]=str1[h]-'0';
    }
    k=b1-b2+1;
    for(i=1;i<=k;i++)
      c=c+a1[i];
      if(c==0)
      printf("0
");
      else {
    while(1)
    {
        if(a1[k]==0)k--;
        else break;
    }
    for(i=k;i>0;i--)
    printf("%d",a1[i]);
    printf("
");}}
    for(i=0;i<11000;i++)
    { str3[i]=0;a1[i]=0;a2[i]=0;}}
return 0;
}
void fun()
{    int i,j,c;
for(;;)
 {for(i=b2-1;i>=0;i--)
      str3[b2-i-1]=a1[i]+'0';
      str3[i]=0;
      if((strcmp(str3,str2)>=0)||a1[b2]>0)
  { 
 for(i=0;i<=b2;i++)
   {
         if(a1[i]<a2[i]) 
       { a1[i]=a1[i]+10-a2[i];a1[i+1]=a1[i+1]-1;
       }
        else a1[i]=a1[i]-a2[i];
    }
 }
   else break;
 }
}