2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15197 Accepted Submission(s): 4695

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

题目不难，就是要知道取模运算的基本法则这题主要是(a*b)%c=(a%c * b%c)%c.

代码：

#include<iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

int main(void)

{

	int n;

	while (cin>>n)

	{

		if(n%2==0||n==1)//n为1或者偶数一定无解

			printf("2^? mod %d = 1\n",n);

		else//奇数一定有解

		{

			int ans=1,t=2;

			while (t%n!=1)

			{

				t=(t*2)%n;//每次都求余数据就不会溢出

				ans++;

			}

			printf("2^%d mod %d = 1\n",ans,n);

		}

	}

	return 0;

}

巴特西

HDU——1395 2^x mod n = 1（取模运算法则）

2^x mod n = 1

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