Educational Codeforces Round 50 (Rated for Div. 2) F

题目链接：http://codeforces.com/contest/1036/problem/F

题意：

题解：求在[2,n]中,x != a ^ b（b >= 2 即为gcd）的个数，那么实际上就是要求x = a ^ b的个数，然后用总数减掉就好了，答案即为。（pow会丢失精度，学习避免精度丢失的方法）

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define LL __int128
#define ull unsigned long long
#define mst(a,b) memset((a),(b),sizeof(a))
#define mp(a,b) make_pair(a,b)
#define pi acos(-1)
#define pii pair<int,int>
#define pb push_back
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MAXN = 1e5 + 10;
const int MAXM = 2e6 + 10;
const ll mod = 1e9 + 7;

bool check[110];
int prime[110];
int mu[110];

void Moblus(int N) {
    mst(check, false);
    mu[1] = 1;
    int tot = 0;
    for(int i = 2; i <= N; i++) {
        if(!check[i]) {
            prime[tot++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < tot; j++) {
            if(i * prime[j] > N) {
                break;
            }
            check[i * prime[j]] = true;
            if(i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            } else {
                mu[i * prime[j]] = -mu[i];
            }
        }
    }
}

int main()
{
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    Moblus(100);
    int t;
    scanf("%d",&t);
    while(t--) {
        ll n;
        scanf("%lld",&n);
        ll ans = 0;
        for(int i = 2; i <= 60; i++) {
            if(!mu[i]) continue;
            ll raiz = round(pow(n, 1.0 / i));
            while(pow((long double)raiz, i) > (long double)n) raiz--;
            while(pow((long double)raiz + 1, i) <= (long double)n) raiz++;
            raiz--;
            ans += (ll)mu[i] * raiz;
        }
        printf("%lld
",n - 1 + ans);
    }
    return 0;
}