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

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 int b1,b2,b3,a1[110000],a2[110000],n;
 4 char str1[110000],str2[110000],str3[110000],*p;
 5 int main()
 6 { 
 7 void fun();
 8 p=str1;
 9     int i,j,k,x=0,c;
10    while( gets(str1))
11    { 
12    for(i=0;i<11000;i++)
13        { if(str1[i]==' ')
14         {str1[i]='\0';
15   strcpy(str2,p+i+1);break;}}
16      
17      
18     b1=strlen(str1);
19     b2=strlen(str2);
20     if(b1<b2){puts(str1); }
21       else if(b1==b2&&(strcmp(str1,str2)<0)){puts(str1);}
22     else
23     {int h=b2-1;
24     for(i=b2-1;i>=0;i--)
25       a1[b2-i-1]=str1[i]-'0';
26     for(i=b2-1;i>=0;i--)
27       a2[b2-i-1]=str2[i]-'0';
28     for(j=0;j<=b1-b2;j++)
29     {
30       fun();
31        h++;
32            for(i=b2;i>0;i--)
33            a1[i]=a1[i-1];
34            a1[0]=str1[h]-'0';
35     }
36     k=b1-b2+1;
37     for(i=1;i<=k;i++)
38       c=c+a1[i];
39       if(c==0)
40       printf("0\n");
41       else {
42     while(1)
43     {
44         if(a1[k]==0)k--;
45         else break;
46     }
47     for(i=k;i>0;i--)
48     printf("%d",a1[i]);
49     printf("\n");}}
50     for(i=0;i<11000;i++)
51     { str3[i]=0;a1[i]=0;a2[i]=0;}}
52 return 0;
53 }
54 void fun()
55 {    int i,j,c;
56 for(;;)
57  {for(i=b2-1;i>=0;i--)
58       str3[b2-i-1]=a1[i]+'0';
59       str3[i]=0;
60       if((strcmp(str3,str2)>=0)||a1[b2]>0)
61   { 
62  for(i=0;i<=b2;i++)
63    {
64          if(a1[i]<a2[i]) 
65        { a1[i]=a1[i]+10-a2[i];a1[i+1]=a1[i+1]-1;
66        }
67         else a1[i]=a1[i]-a2[i];
68     }
69  }
70    else break;
71  }
72 }

View Code

 

标签：110000,str3,Big,str1,Number,a1,b2,than
来源： https://www.cnblogs.com/gznb/p/11213703.html