LightOJ1220 Mysterious Bacteria

题目链接:点我

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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

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题意:

给你一个公式 x=bp, 给你一个数x,求最大的p.

思路:

唯一分解定理:x = pe11 * pe22 *pe33 ….,那么根据题意所有ans = gcd( e1 ,e2,e3…….);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;}