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hdoj-1395-2^x mod n = 1【欧拉定理】

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2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14349 Accepted Submission(s): 4438


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

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#include<stdio.h>
int main(){
	int n;
	while(~scanf("%d",&n)){
		if(n==1||n%2==0) printf("2^? mod %d = 1\n",n);
		else{
			int k=2,ncas=1;
		    while(1){
		    	if(k%n==1) break;
		    	k=k%n*2;
		    	++ncas;
		    }
		    printf("2^%d mod %d = 1\n",ncas,n);
		} 
	}
	return 0;
}


欧拉定理 :
对于互质的正整数 a 和 n ,有 a ^φ(n) ≡ 1 mod n 。 φ(n) 指小于n且与n互质的的正整数(包括1)的个数;
因为2与所有奇数都互质,所以对与奇数n,总会存在x,使得:  2^x =1(mod n)

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hdoj-1395-2^x mod n = 1【欧拉定理】

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原文地址:http://blog.csdn.net/qq_18062811/article/details/47301609

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