HDU 5297

用x ^ （1 / n） 来求个数，容斥原理 ， Num会向后移动， 迭代到不再变化，退出循环

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
const int Primes[] = {-2,-3,-5,-7,-11,-13,-17,-19,-23,-29,-31,-37,-41,-43,-47,-53,-59,-61,-67,-71};
vector<int> R_number;
void GetR(LL R) {
    R_number.clear();
    for(int i = 0; abs(Primes[i])<= R; ++i) {
        int tmp = Primes[i];
        int Num = R_number.size();
        for(int j = 0; j < Num; ++j) {
            if(abs(R_number[j]*tmp) <= 63)
            R_number.push_back(tmp * R_number[j]);
        }
        R_number.push_back(Primes[i]);
    }
}
LL Cnt(LL N) {
    //GetR(R);
    if(N == 1) return 0;
    LL cnt = N;
    for(int i = 0; i < R_number.size(); ++i) {
        LL tmp = LL (pow(N+0.5, 1.0/abs(R_number[i])) - 1);
        if(R_number[i] < 0) cnt -= tmp;
        else cnt += tmp;
    }
    return cnt - 1;
}
void Solve(LL N,LL R) {
    GetR(R);
    LL ans = N;
    while(1) {
        LL tmp = Cnt(ans);
        if(tmp == N) break;
        ans += N - tmp;
    }
    printf("%lld\n",ans);
}
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        LL n,r;
        scanf("%I64d %I64d",&n,&r);
        Solve(n,r);
    }
    return 0;
}