HDU 2685 I won't tell you this is about number theory

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2685

题意：求gcd(a^m - 1, a^n - 1) mod k

思路：gcd(a^m - 1, a^n - 1) = a^gcd(m, n) - 1

code：

 1 #include <stdio.h>
 2 
 3 int gcd(int a, int b)
 4 {
 5     return !b ? a : gcd(b, a%b);
 6 }
 7 
 8 int mod_pow(int a, int x, int mod)
 9 {
10     int tmp = a;
11     int ret = 1;
12     while(x) {
13         if (x & 1) {
14             ret = ret * tmp % mod;
15         }
16         tmp = tmp * tmp % mod;
17         x >>= 1;
18     }
19 
20     return ret;
21 }
22 
23 int main()
24 {
25     int t, a, m, n, k;
26     scanf("%d", &t);
27     while (t--) {
28         scanf("%d %d %d %d", &a, &m, &n, &k);
29         int d = gcd(m, n);
30         int ans = mod_pow(a, d, k);
31         printf("%d\n", (ans - 1 + k) % k); 
32     }
33 
34     return 0;
35 }

 

posted @ 2016-04-19 20:17  jasaiq  阅读(300)  评论(0)  编辑  收藏  举报