hdoj 1359 2^x mod n = 1

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本文介绍了一个通过预计算解决2^xmodn=1问题的方法，寻找满足条件的最小正整数x。利用双重循环遍历不同n值下2的幂次方模n等于1的情况，并存储结果用于快速查询。

http://acm.hdu.edu.cn/showproblem.php?pid=1395

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11640 Accepted Submission(s): 3624

Problem Description

Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.

Input

One positive integer on each line, the value of n.

Output

If the minimum x exists, print a line with 2^x mod n = 1.

Print 2^? mod n = 1 otherwise.

You should replace x and n with specific numbers.

Sample Input

2
5

Sample Output

  
   2^? mod 2 = 1
2^4 mod 5 = 1

#include<stdio.h>
#include<string.h>
int a[10000];
int main()
{
	int i,j,k;
	memset(a,0,sizeof(a));
	for(i=1;i<10000;i++)
	 for(j=1,k=2;j<10000;j++,k=k*2)
	 {
	 	k=k%i;
	    if(k%i==1)
	    {
	 	   a[i]=j;
	 	   break;
	    }
	}
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		if(a[n]==0)
		printf("2^? mod %d = 1\n",n);//注意格式，一直提交都没加 = 1 
		else 
		printf("2^%d mod %d = 1\n",a[n],n);
	}
	return 0;
}