HDOJ 题目1395 2^x mod n = 1（水题 易错）

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11712    Accepted Submission(s): 3649

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

25

 

Sample Output

2^? mod 2 = 12^4 mod 5 = 1

 

Author

MA, Xiao

 

Source

ZOJ Monthly, February 2003

 

ac代码

#include<stdio.h>int main(){int n;while(scanf("%d",&n)!=EOF){if(n%2==0||n==1)//注意没有满足的情况下的条件，n==1，不要忘了{printf("2^? mod %d = 1\n",n);}else{int s=1,d=0;while(1){s*=2;d++;s%=n;//注意这儿，应用剩余定理，去掉会超时if(s%n==1){printf("2^%d mod %d = 1\n",d,n);break;}}}}}