HDU1395 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): 7004    Accepted Submission(s): 2106

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

 

//代码一：-----TLE
#include<cstdio>
#include<cstring>
const int MAX=10000000;
bool flag[MAX];

int main()
{
	int n,i,k;
	while(~scanf("%d",&n))
	{
		if((n&1)==0)
			printf("2^? mod 2 = 1\n");
		else
		{
			memset(flag,false,sizeof(flag));
			for(i=1,k=2;;++i)
			{
				if(k%n==1)
				{
					printf("2^%d mod %d = 1\n",i,n);
					break;
				}
				else if(flag[k%n])
				{
					printf("2^? mod 2 = 1\n");
					break;
				}
				else
					flag[k%n]=true;
				k<<=1;
			}
		}
	}
	return 0;
}

//代码二：  copy代码：除了1和偶数不行，其他都是能找到的(but why????)
int main(void)
{
    int n;
    while(~scanf("%d",&n))
	{
        if(n == 1 || n % 2 == 0)
		{
            printf("2^? mod %d = 1\n", n);
            continue;
        }
        int k = 1, ans = 2;
        while(ans != 1)
		{
            ans = ans *2 % n;
            k++;
        }
        printf("2^%d mod %d = 1\n", k, n);
    }
    
    return 0;
}

//代码三：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
bool h[10000];
int main()
{
    int n,t,k;
    while(scanf("%d",&n)!=EOF)
    {
        if(n%2==0||n==1)
        {
			printf("2^? mod %d = 1\n",n);
			continue;
		}
        memset(h,false,n*sizeof(bool));
        k=1;
		t=2;
		h[2]=1;
        while(t%n!=1)
        {
            k++;
            t<<=1;
            t=t%n;
            if(h[t]) 
				break;
            h[t]=true;
        }
        if(t%n!=1)
          printf("2^? mod %d = 1\n",n);
        else
          printf("2^%d mod %d = 1\n",k,n);
    }
    return 0;
}