题目地址： http://acm.hdu.edu.cn/showproblem.php?pid=1395
题目描述：

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12685    Accepted Submission(s): 3951


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
题意： 求最小的指数是公式成立。 题解： 同余定理的运用，可以认为 t 是等于 t%n的，如果涉及到乘法然后取余等运用同余的地方。 (a * b) % n = ((a % n) * b) % n        (同余同取余) 代码： #include #define MM 10000 long long two[MM] = {0}; int x=0,n=0; int main() { while(scanf("%d",&n)!=EOF) { int t = 1; for(x=1;x<=MM-1;x++) { t*=2; if(t%n==1) break; else t=t%n;//sub the number 's range } if(x