HDU1395 ZOJ1489 2^x mod n = 1【暴力法＋数论】

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17824    Accepted Submission(s): 5582

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

 

Author

MA, Xiao

 

Source

ZOJ Monthly, February 2003

问题链接：HDU1395 ZOJ1489 2^x mod n = 1

问题简述：对于输入的n，求满足 2^x mod n = 1的最小正整数x。

问题分析：若n是偶数或者1则无解，否则用暴力法来解。这个题是一个水题。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* HDU1395 ZOJ1489 2^x mod n = 1 */

#include <iostream>
#include <stdio.h>

using namespace std;

int main()
{
    int n;
    while(scanf("%d", &n) != EOF) {
        if(n % 2 == 0 || n == 1)
            printf("2^? mod %d = 1\n", n);
        else {
            int t = 2, x = 1;
            while(t != 1) {
                t *= 2;
                t %= n;
                x++;
            }
            printf("2^%d mod %d = 1\n", x, n);
        }
    }

    return 0;
}