HDU1395_2^x mod n = 1【数论】【水题】

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12605    Accepted Submission(s): 3926

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

题目大意：给你一个数N，判断是否存在x，满足2^x mod N = 1。若

满足，对于满足条件的最小x，输出2^x mod N = 1，否则输出

2^? mod 2 = 1。

思路：用到数论上的乘法逆元的规律了。

乘法逆元：对于整数a、p如果存在整数b，满足a*b mod p = 1，则称

b是a的模p的乘法逆元。a存在模p的乘法逆元的充要条件是gcd(a,p) = 1

此题中，令a = 2^x，b = 1，p = n，则若存在x使得2^x mod N = 1，

则gcd(2^x,N) = 1。

1>.因为N>0，当N为偶数时，gcd(2^x,N) = 2*k(k=1,2,3……)，不满足

2>.当N为奇数时，gcd(2^x,N) = 1满足条件。

3>.当N为1时，2^x mod N = 0，不符合条件

所以N为奇数，且不为1，满足2^x mod N = 1，暴力求解。

参考博文：http://blog.csdn.net/mxway/article/details/8954364

#include<stdio.h>

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

    return 0;
}