F. Relatively Prime Powers

题意：

给你一个n，问满足在[2,n]范围内有多少个数是非次方数（也就是不是 $x^k(k>1)$ 这样的）

思路：

答案就是

原理是利用容斥，注意n开i次根是向下取整（这题巨卡精度）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
#define LL long long
#define LD long double
#define mod 1000000007
int cnt, flag[105] = {1,1}, pri[105], mu[105] = {0,1};
LL sum[105];
void Mulset()
{int i, j;for(i=2;i<=105;i++){if(flag[i]==0){pri[++cnt] = i;mu[i] = -1;}for(j=1;j<=cnt&&i*pri[j]<=105;j++){flag[i*pri[j]] = 1;if(i%pri[j]==0){mu[i*pri[j]] = 0;break;}mu[i*pri[j]] = -mu[i];}}
}
int main(void)
{LL ans, n;int T, i;Mulset();scanf("%d", &T);while(T--){ans = 0;scanf("%lld", &n);for(i=2;i<=60;i++){sum[i] = (LL)powl((LD)n+0.1, (LD)1.0/i)-1;ans += sum[i]*mu[i];}printf("%lld\n", n-1+ans);}return 0;
}

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Educational Codeforces Round 50: F. Relatively Prime Powers（莫比乌斯函数）

F. Relatively Prime Powers

题意：

思路：

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