问题及代码:
运行结果:
/*
*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:单昕昕
*完成日期:2015年2月3日
*版 本 号:v1.0
*
*问题描述: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.
*程序输入: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.
*程序输出: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<iostream>
#include<cstring>
using namespace std;
int main()
{
char str[100000];
int s,len,sum,i;
while(cin>>str>>s)
{
len=strlen(str);
sum=0;
for(i=0;i<len;i++)
sum=(sum*10+(str[i]-'0')%s)%s;
cout<<sum<<endl;
}
return 0;
}
运行结果:
大数取模。
我一开始想会不会有下面这些的结论,后来一百度发现还真有。
对于大数问题,从来都只能巧胜~~~
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
学习心得:
大数问题真是硬伤,上周的题目被虐的死去活来的。。