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hdu 1395 2^x mod n = 1(暴力题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1395
2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12146 Accepted Submission(s): 3797
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
题目大意:暴力搜索,找到合适的X值,这一题可以采取反过来暴力寻找,这一简单易懂些。
要注意的是输出的值时都要变化的,输出注意一下就好了,毕竟我是wa过的。。。
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 int main () 6 { 7 int n; 8 while (cin>>n) 9 {10 if (n%2&&n>1)11 {12 int s=1,x=1;13 while (x)14 {15 s=s*2%n;16 if (s==1)17 {18 printf ("2^%d mod %d = 1\n",x,n);19 break;20 }21 x++;22 }23 }24 else25 printf ("2^? mod %d = 1\n",n);26 }27 return 0;28 }
hdu 1395 2^x mod n = 1(暴力题)
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