Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7329    Accepted Submission(s): 5060

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

2 312 7152455856554521 3250

 

Sample Output

251521
大数取余现在遇到了两种，一种是A^BmodC，这种用快速幂求解，有固定的模板，而现在这种是大数Amod B,B是int型，A是个大数，可能无法直接存储，可以用字符串来保存，可以设置变量t从大数的第一位向后遍历赋值，如果大于mod就取余更新，就像数学计算一样。
例：123mod6，取中间变量t，1<6,t=1;
-2t=1乘10+2%6=0 t=0;
-3t=0乘10+3%6=3;t=3;
<span style="font-size:18px;">#include<cstdio>#include<cstring>int main(){long long b,i,s,h,j;char a[1100];while(scanf("%s%lld",a,&b)!=EOF){s=0;j=0;h=strlen(a);for(i=0;i<h;i++){s=s*10+(a[i]-'0');s%=b;}printf("%lld\n",s);}return 0;}</span>