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): 11563    Accepted Submission(s): 3588

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​​

思路：大致题意：给定一个 n 值, 求出满足式: 2^x mod n = 1 的最小 x 值;

1): 2^n 为偶数, n 为偶数就直接 pass 了, n 为奇数时肯定有结果(退出循环条件)
2): 两个容易 TLE 的地方:
  只判断了 n 是否为偶数, 没有考虑到 n <= 1的情况(因为 x > 0)
  在累乘第一个操作数时, 没有取模运算, 导致值太大, TLE; 第一个操作数2^n
取模理解: (a * b) % n = (a % n * b % n) % n;

理解就能做出了。

/******************
Author:jiabenmuwei
sources:HDU1395
times:15ms
******************/
#include<stdio.h>
int main()
{

    int n,i,j;
    while(~scanf("%d",&n))
    {
        if(n%2==0||n<=1)
        {
            printf("2^? mod %d = 1\n",n);
            continue;
        }
        int a=1;
        for(i=1;;i++)
        {
            a*=2;
            if(a%n==1)break;
            a%=n;

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


}

C++:

#include<iostream>
using namespace std;
int main()
{
    int i,n,a;
    while(scanf("%d",&n)!=EOF)
    {
        if(n%2==0||n<=1)
        {
            printf("2^? mod %d = 1\n",n);
            continue;
        }
        a=1;
        for(i=1;; i++)
        {
            a*=2;
            if(a%n==1)break;
            a%=n;
        }
        printf("2^%d mod %d = 1\n",i,n);
    }
    return 0;
}