LightOJ 1220 Mysterious Bacteria

题目链接:点我

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

317107374182425

Sample Output

Case 1: 1Case 2: 30Case 3: 2

题意:

给你一个公式 x=bpx = b ^{p}, 给你一个数x,求最大的p.

思路:

唯一分解定理:x = pe11p_1^{e_1} * pe22p_2^{e_2} *pe33p_3^{e_3} ….,那么根据题意所有ans = gcd( e1e_1 ,e2e_2,e3e_3…….);而且这里,如果x为负数的话,那么ans一定只能是奇数.
代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
using namespace std;typedef long long LL;
const int maxn = 1e6 + 10;
bool vis[maxn];
int pri[maxn/10];
int k , num = 0;
LL n, ans;
bool flag;
LL a[1000];void prime(){memset(vis,false,sizeof(vis));for(int i = 2; i < maxn; ++i){if (!vis[i]) pri[++num] = i;for(int j = 1;j <= num && i *pri[j] < maxn; ++j){vis[i * pri[j]] = true;if (i % pri[j] == 0) break;}}
}LL gcd(LL a, LL b){return b ? gcd(b, a % b) : a;
}void solve(){for(int i = 1; i <= num && pri[i] <= (int)sqrt(n); ++i){int cnt = 0;while(n % pri[i] == 0){++cnt;n /= pri[i];}if(cnt) a[++k] = cnt;if (cnt == 1){ans = 1;return ;}} if(n > 1) {ans = 1;return ;}int tmp = a[1];for(int i = 2; i <= k; ++i){tmp = gcd( tmp, a[i]);}if(flag){//如果n为奇数,那么ans 只能为奇数,while(tmp % 2 ==0)tmp /= 2;}ans = tmp;
}int main(){int t, kase = 0;prime();scanf("%d", &t);while( t--){scanf("%lld", &n);flag = false;if(n < 0){flag = true;n = -n;}k = 0;ans = 0;solve();printf("Case %d: %lld\n", ++kase, ans);}return 0;
}

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