Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9006 Accepted Submission(s): 6100
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
(ABC)%n=(A*100%n+B*10%n+C%n)%n
(A*B)%n=(A%n*B%n)%n
于是，可以利用循环，对于挺大的数，利用字符串来表示这个数。
比如，令ABC为字符串，则有
1)sum=A%n;
2)A%n*10+B%n=sum+B%n
于是sum=(sum+B%n)%n
3)经过循环，会有：
sum=(sum+C%n)%n=((sum+B%n)%n*10+C%n)%n=((A%n*10+B%n)%n*10+C%n)%n=((A*100%n+B*10%n)%n+C%n)%n=(A*100%n+B*10%n+C%n)%n
故而可以用循环求解。
#include <iostream>
#include <string>
using namespace std;
int main()
{
string a;
int b,len,sum;
while(cin>>a>>b)
{
len=a.length();
sum=;
for(int i=;i<len;i++)
sum=(sum*+(a[i]-'')%b)%b;
cout<<sum<<endl;
}
return ;
}
用JAVA更简单
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
int n;
BigInteger a,b;
while(in.hasNextBigInteger()) {
a=in.nextBigInteger();
b=in.nextBigInteger();
a=a.remainder(b);
System.out.println(a);
}
}
}