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=0; for(int i=0;i<len;i++) sum=(sum*10+(a[i]-'0')%b)%b; cout<<sum<<endl; } return 0; } View Code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); } } } View Code
转载于:https://www.cnblogs.com/Weixu-Liu/p/9165366.html
相关资源：JAVA上百实例源码以及开源项目