2019 Hang electric multi-school fourth hdu6623 Minimal Power of Prime

Minimal Power of Prime

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Problem-solving ideas

Hit \ (N ^ \ frac {1 } {5} \) prime number table therein, for each n, the first decomposed \ (N ^ \ frac {1 } {5} \) prime numbers in the range of decomposition after n to m, if m is equal to 1, the answer is \ (N ^ \ frac {1 } {5} \) primes decomposed inside the minimum number k. Otherwise, continue to break down, then prime numbers are used to break greater than \ (N ^ \ frac {1 } {5} \) , so a maximum of four prime numbers multiplied, so only three cases: \ (^ P . 4 \) , \ (P ^. 3 \) , \ (P ^ 2 \) , \ (P ^ 2 * Q ^ 2 \) , and the answer is the case (P, Q 1 is more than \ (N ^ \ frac 5} {1} {\) primes). For the first two cases, respectively, see \ (m ^ \ FRAC. 1 {{}}. 4 \) , \ (m ^ \ FRAC. 1 {{}}. 3 \) is a number, for a three-four cases, in fact, is the same as long as the look \ (m ^ \ frac {1 } {2} \) is not an integer on the line, the first four of which are not, the answer is 1.

code show as below

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

ll p(ll a, int b)
{
    ll ans = 1;
    for(int i = 1; i <= b; i ++)
        ans *= a;
    return ans;
}

int main()
{
    int t;
    t = read();
    vector<int> vec;
    for(int i = 2; i <= 4000; i ++){
        bool flg = true;
        for(int j = 2; j <= sqrt(i); j ++){
            if(i % j == 0){
                flg = false;
                break;
            }
        }
        if(flg)
            vec.push_back(i);
    }
    while(t --){
        ll n;
        scanf("%lld", &n);
        int ans = 100;
        for(int i = 0; i < vec.size(); i ++){
            int t = vec[i];
            int cnt = 0;
            while(n % t == 0){
                n /= t;
                ++cnt;
            }
            if(cnt)
                ans = min(ans, cnt);
        }
        if(n == 1)
            printf("%d\n", ans);
        else {
            for(int i = 4; i >= 1; i --){
                if(i == 1)
                    printf("1\n");
                else {
                    ll x = max((ll)pow(n, 1.0/i) - 5, 1LL);
                    bool flg = true;
                    ll m = p(x, i);
                    while(m < n){
                        ++ x;
                        m = p(x, i);
                    }
                    if(m == n){
                        printf("%d\n", min(ans, i));
                        break;
                    }
                }
            }
        }
    }
    return 0;
}