UVa 10622 (gcd 分解质因数) Perfect P-th Powers

 1 #include <cstdio>
 2 #include <cmath>
 3 
 4 const int maxn = 46500;
 5 bool vis[maxn + 10];
 6 int prime[4810], cnt = 1;
 7 
 8 void Init()
 9 {
10     int m = sqrt(maxn + 0.5);
11     for(int i = 2; i <= m; ++i) if(!vis[i])
12         for(int j = i * i; j <= maxn; j += i) vis[j] = true;
13     for(int i = 2; i <= maxn; ++i) if(!vis[i]) prime[cnt++] = i;
14     cnt--;
15 }
16 
17 int gcd(int a, int b)
18 {
19     return b == 0 ? a : gcd(b, a % b);
20 }
21 
22 int solve(int n)
23 {
24     int ans = 0;
25     for(int i = 1; i <= cnt; ++i)
26     {
27         int k = 0;
28         while(n % prime[i] == 0)
29         {
30             k++;
31             n /= prime[i];
32         }
33         if(k)
34         {
35             if(k == 1) return 1;
36             ans = gcd(k, ans);
37         }
38     }
39     if(n > 1) return 1;
40     return ans;
41 }
42 
43 int main()
44 {
45     Init();
46 
47     int n;
48     while(scanf("%d", &n) == 1 && n)
49     {
50         int ans = solve(n < 0 ? -n : n);
51         if(n < 0) while((ans & 1) == 0) ans >>= 1;
52         printf("%d\n", ans);
53     }
54 
55     return 0;
56 }