HDU 1212 Big Number 【大数】

发布时间：2020-12-14 02:27:16 所属栏目：大数据来源：网络整理

导读：Big Number Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5984????Accepted Submission(s): 4182 Problem Description As we know,Big Number is always troublesome. But it's really impor

Big Number

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5984????Accepted Submission(s): 4182

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

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

思路：

???? 对大数取余，需要将大数（原来在字符串中的）转换到整型数组里面，从最高位开始对正整数c取余，最终输出c的余数就行了！

代码：

#include <stdio.h>
#include <string.h>
#define N 1005
char a[N];
int b[N];
int main()
{
	int len,n,i,j,k,t,s,c;
	while(scanf("%s%d",a,&c)!=EOF)
	{
		memset(b,sizeof(b));
		len=strlen(a);
	    for(i=0;i<len;i++)//将位将字符转换成数字，保存到整形数组里面！（注：a[0]保存的是最高位！） 
	       b[i]=a[i]-'0';
	    for(i=0,t=0;i<len;i++)	
	    {
	    	
	          t=t*10+b[i];	
	          t=t%c;//从最高位开始逐位对c取余！ 
	    }
	    printf("%dn",t);
	}
	return 0;
}

（编辑：李大同）

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