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Description
Dr. Mob has just discovered a Deathly Bacteria. He named it RC-01. RC-01 has a very strange reproduction system. RC-01 lives exactly x days. Now RC-01 produces exactly p new deadly Bacteria where x = bp (where b, p are integers). More generally, x is a perfect pth power. Given the lifetime x of a mother RC-01 you are to determine the maximum number of new RC-01 which can be produced by the mother RC-01.
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
Each case starts with a line containing an integer x. You can assume that x will have magnitude at least 2 and be within the range of a 32 bit signed integer.
Output
For each case, print the case number and the largest integer p such that x is a perfect pth power.
Sample Input
3
17
1073741824
25
Sample Output
Case 1: 1
Case 2: 30
Case 3: 2
题意：给定x，找出最大的p使x=b^p；
思路：素数分解。x=(p1^k1)*(p2^k2)*...*(pn^kn) = (p1^a1*p2^a2*...*pn^an)^p=b^p，即提出来一个最大的p，p=gcd(k1,k2,...,kn)。
注意点：素数分解时由于素数表只能到10^7，所以需要对最后一个因子进行特判。
坑点：x为负数的时候p不能为偶数，这时只要把之前提出来的所有2都乘进去即可，剩下留在外面的奇数p就是答案了。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> using namespace std; typedef long long ll; const int maxn=10001000; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); ll x,p; bool isprime[maxn]; vector<ll> prime; void play_prime() { memset(isprime,1,sizeof(isprime)); isprime[1]=0; for(int i=2;i<maxn;i++){ if(!isprime[i]) continue; for(int j=i+i;j<maxn;j+=i){ isprime[j]=0; } } for(int i=1;i<maxn;i++){ if(isprime[i]) prime.push_back(i); } } int main() { int T;cin>>T; play_prime(); int tag=1; while(T--){ cin>>x; bool f=0; if(x<0) f=1,x=-x; p=-1; bool flag=1; for(int i=0;i<(int)prime.size()&&prime[i]*prime[i]<=x;i++){ if(x%prime[i]) continue; ll k=0; while(x%prime[i]==0){ x/=prime[i]; k++; } if(p==-1) p=k; else p=__gcd(p,k); } if(x!=1) p=1; if(f){ while(p%2==0) p/=2; } cout<<"Case "<<tag++<<": "<<p<<endl; } return 0; }
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原文地址：http://www.cnblogs.com/--560/p/4564430.html
