Perfect Pth Powers
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16383 | Accepted: 3712 |
Description
We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More generally, x is a perfect pth power if, for some integer b, x = bp. Given an integer x you are to determine the largest p such that x is a perfect pth power.Input
Each test case is given by a line of input containing x. The value of x will have magnitude at least 2 and be within the range of a (32-bit) int in C, C++, and Java. A line containing 0 follows the last test case.
Output
For each test case, output a line giving the largest integer p such that x is a perfect pth power.
Sample Input
17 1073741824 25 0
Sample Output
1 30 2
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <math.h> 5 #define MAXN 100010 6 #define INF 0x7fffffff 7 using namespace std; 8 const int N = 77000 ; 9 int np[N >> 6], p[N >> 2],ans[N]= {0}; 10 void init() 11 { 12 int n = N; 13 p[0] = 1, p[1] = 2; 14 for (int i = 3; i <= n; i++, i++) 15 { 16 if (np[i >> 6] & (1 << ((i >> 1) & 31))) continue; 17 p[++p[0]] = i; 18 for (int j = (i << 1) + i; j <= n; j += (i << 1)) 19 { 20 np[j >> 6] |= (1 << ((j >> 1) & 31)); 21 } 22 } 23 } 24 int gcd(int x,int y) 25 { 26 if(y==0)return x; 27 return gcd(y,x%y); 28 } 29 int fun(long long x) 30 { 31 int i,ans=0,an=0; 32 bool flag=0; 33 if(x<0)flag=1,x=-x;; 34 for(i=1; x>=p[i]&&i<=p[0]; i++) 35 { 36 if(x%p[i]==0) 37 { 38 ans=0; 39 while(x%p[i]==0)ans++,x/=p[i]; 40 if(an==0)an=ans; 41 else 42 { 43 an=gcd(ans,an); 44 } 45 } 46 } 47 if(x!=1) 48 { 49 an=1; 50 } 51 while(flag&&(an%2==0))an>>=1; 52 if(an==0)return 1; 53 return an; 54 } 55 int main() 56 { 57 init(); 58 long long n; 59 while(~scanf("%I64d",&n),n) 60 { 61 printf("%d ",fun(n)); 62 } 63 }