HDU1395 ZOJ1489 2^x mod n = 1【暴力法+數論】
阿新 • • 發佈:2019-02-17
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17824 Accepted Submission(s): 5582
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
Total Submission(s): 17824 Accepted Submission(s): 5582
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
問題簡述 :對於輸入的n,求滿足 2^x mod n = 1的最小正整數x。
問題分析:若n是偶數或者1則無解,否則用暴力法來解。這個題是一個水題。
程式說明:(略)
題記:(略)參考連結:(略)
AC的C++語言程式如下:
/* HDU1395 ZOJ1489 2^x mod n = 1 */ #include <iostream> #include <stdio.h> using namespace std; int main() { int n; while(scanf("%d", &n) != EOF) { if(n % 2 == 0 || n == 1) printf("2^? mod %d = 1\n", n); else { int t = 2, x = 1; while(t != 1) { t *= 2; t %= n; x++; } printf("2^%d mod %d = 1\n", x, n); } } return 0; }