hdu 1212 Big Number（大数的模）

小咪咪 2022-06-08 02:57 140阅读 0赞

# Big Number #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 8449    Accepted Submission(s): 5832**

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
      
       
      
       
       
        题目大意：求一个长度不超过1000的数mod B
      
       
      
       
       
        思路：比如说一个数4649874564651215462 mod 1000，怎么做呢？ 一位一位的mod，先用4mod1000，得到的4，用得到的值乘以10再加上下一个数6，用46mod1000，得到46，再乘以10家上下一个数4，464.........一直下去，4649mdo1000，得到649...... 一直到数组的最后一位。
      
       
      
       
       
        AC代码：
      
       
      
       
    
      
       
      
       
       
        #include<cstdio>
    #include<cstring>
    using namespace std;
    int main()
    {
    	char a[1005];
    	int b;
    	while(scanf("%s %d", a, &b) != EOF)
    	{
    		int answer;
    		int length = strlen(a);
    		for(int i = 0; i < length; i++){
    			if(i == 0)
    				answer = (a[i] - 48) % b;  //将char变成int型，要减去48 
    			else
    				answer = (answer * 10 + (a[i] - 48)) % b;
    		}
    		printf("%d\n", answer);
    	}
    	return 0;	
    }