HDU 1395 2^x mod n = 1

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17023    Accepted Submission(s): 5313

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

 

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直接暴力，偶数和1不可能，其余奇数都可以

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<cmath>
using namespace std;
#define maxn 1010
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n%2==0||n==1)
        {
            printf("2^? mod %d = 1
",n);
            continue;
        }
        int m = 1,num;
        for(int i=1; ;i++)//因为绝对可以，所以这里写成死循环，暴力
        {
            m = m * 2;
            m = m%n;//注意这里要取余！！！
            if(m%n==1)
            {
                num = i;
                break;
            }

        }
        printf("2^%d mod %d = 1
",num,n);
    }
    return 0;
}