HDU 1395
2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16883 Accepted Submission(s): 5262
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
#include "stdio.h" int main() { int n; while(~scanf("%d",&n)&&n){ if(n==1||n%2==0){ printf("2^? mod %d = 1\n",n); continue; } else{ int x=1,mut=2; mut%=n; while(mut!=1){ x++; mut=mut*2%n; } printf("2^%d mod %d = 1\n",x,n); } } return 0; }
开始时Time Limit Exceeded
改进1.n==1情况 2.循环式取余,减少运算量