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
题目链接
对每个数 X有 X=aY,求 Y的最大值。
对于每一个 X进行质因数分解
X=p1e1×p2e2×...×pkek
∴ 当 Y=gcd(e1,e2,...,ek)时可以将 p合并为一个数 M
X=Mgcd(e1,e2,...,ek)
此时 Y具有最大值。
这道题目 X并不是正整数,会有负数,所以当 X为负数时需要将其转换为正数求解
当 X为负数且 Y最大值为偶数(显然不能为偶数)时,需要将 M逐次乘方直到 Y为奇数。
AC代码:
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define XDebug(x) cout << #x << "=" << x << endl;
#define ArrayDebug(x,i) cout << #x << "[" << i << "]=" << x[i] << endl;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e7 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
char c;
int sgn;
if (c = getchar(), c == EOF) {
return 0;
}
while (c != '-' && (c < '0' || c > '9')) {
c = getchar();
}
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0');
}
ret *= sgn;
return 1;
}
template <class T>
inline void out(T x) {
if (x > 9) {
out(x / 10);
}
putchar(x % 10 + '0');
}
bool IsPrime[maxn];
ll Prime[maxn / 10];
int tot;
vector<ll> ResolveAns;
inline ll gcd(ll x, ll y) {
return y ? gcd(y, x % y) : x;
}
void PrimeInit() {
mem(IsPrime, 1);
IsPrime[1] = 0;
tot = 0;
for (ll i = 2; i < maxn; ++i) {
if (IsPrime[i]) {
Prime[tot++] = i;
for (ll j = i * i; j < maxn; j += i) {
IsPrime[j] = 0;
}
}
}
}
void Resolve(ll x) {
ResolveAns.clear();
int num = 0;
while (x > 1 && Prime[num] * Prime[num] <= x && num < tot) {
if (!(x % Prime[num])) {
int cnt = 0;
while (!(x % Prime[num])) {
x /= Prime[num];
cnt++;
}
ResolveAns.pb(cnt);
}
num++;
}
if (x > 1) {
ResolveAns.pb(1);
}
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
PrimeInit();
ll T;
read(T);
for (ll Case = 1, N; Case <= T; ++Case) {
bool symbol = 0;
read(N);
if (N < 0) {
N = -N;
symbol = 1;
}
Resolve(N);
ll ans = ResolveAns[0];
for (int i = 1; i < int(ResolveAns.size()); ++i) {
ans = gcd(ans, ResolveAns[i]);
}
if (symbol) {
while (!(ans % 2)) {
ans >>= 1;
}
}
printf("Case %lld: %lld\n", Case, ans);
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}