题目及代码
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
/*
*Copyright (c)2014,烟台大学计算机与控制工程学院
*All rights reserved.
*文件名称：mod.cpp
*作 者：冷基栋
*完成日期：2015年2月20日
*版 本 号：v1.0
*/
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char str[100000];
int s,l,b,i;
while(cin>>str>>s)
{
l=strlen(str);
b=0;
for(i=0;i<l;i++)
b=(b*10+(str[i]-'0')%s)%s;
cout<<b<<endl;
}
return 0;
}
运行结果：
知识点总结：
在做题之前，先了解这样一些结论：
A*B % C = (A%C * B%C)%C
(A+B)%C = (A%C + B%C)%C
如 532 mod 7 =（500%7+30%7+2%7)%7;
当然还有a*b mod c=(a mod c+b mod c)mod c;
如35 mod 3=((5%3)*(7%3))%3
有了这一些结论，题目就好做了！
学习心得：
好好学习 天天向上
杭电ACM 三 大数取模
原文：http://blog.csdn.net/ljd939952281/article/details/43888675
