标签：hduj java
Big Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4622 Accepted Submission(s): 3198
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
JAVA;
import java.util.Scanner;
import java.math.BigInteger;
public class yhrtr {
public static void main(String[] args) {
Scanner cin = new Scanner (System.in);
BigInteger a,b,c;
while(cin.hasNext())
{
a = cin.nextBigInteger();
b = cin.nextBigInteger();
c = a.mod(b);
System.out.println(c);
}
}
}
C++;
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
char c[1005];
int m;
while(cin>>c>>m)
{
int l=strlen(c);
__int64 s=0;
for(int i=0;i<l;i++)
{
s=(s*10+c[i]-'0')%m;
}
printf("%I64d\n",s);
}
return 0;
}
HDUJ 1212 Big Number,布布扣,bubuko.com
HDUJ 1212 Big Number
标签：hduj java
原文地址：http://blog.csdn.net/hyccfy/article/details/38232111
