HDU 1395 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): 17023 Accepted Submission(s): 5313
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.
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
Recommend
直接暴力，偶数和1不可能，其余奇数都可以
#include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<cmath> using namespace std; #define maxn 1010 int main() { int n; while(~scanf("%d",&n)) { if(n%2==0||n==1) { printf("2^? mod %d = 1 ",n); continue; } int m = 1,num; for(int i=1; ;i++)//因为绝对可以，所以这里写成死循环，暴力 { m = m * 2; m = m%n;//注意这里要取余！！！ if(m%n==1) { num = i; break; } } printf("2^%d mod %d = 1 ",num,n); } return 0; }