高精度运算——A mod B
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.
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
仍然是小学知识给的启示。列出一个除法的竖式，从最高位开始除，余数乘10加下一位上的数，再除。循环进行。
所以代码非常简单。
- #include<stdio.h>
- #include<string.h>
- int main()
- {
- intn,sum,i,k;
- char s[1001];
- while(scanf("%s%d",s,&n)!=EOF)
- {
- k=strlen(s);
- sum=0;
- for(i=0;i<k;i++)
- {
- sum=sum*10+s[i]-'0';
- sum=sum%n;
- }
- printf("%d\n",sum);
- }
- return 0;
- }