实际上就是求在[2,n]中,x != a^b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。
直接开方求和显然会有重复的数。容斥搞一下，但实际上是要用到莫比乌斯函数的，另外要注意减掉1^b这种情况。
#include <cmath> #include <cstdio> #include <cstdlib> #include <cassert> #include <cstring> #include <set> #include <map> #include <list> #include <queue> #include <string> #include <iostream> #include <algorithm> #include <functional> #include <stack> using namespace std; typedef long long ll; #define T int t_;Read(t_);while(t_--) #define dight(chr) (chr>='0'&&chr<='9') #define alpha(chr) (chr>='a'&&chr<='z') #define INF (0x3f3f3f3f) #define maxn (300005) #define maxm (10005) #define mod 1000000007 #define ull unsigned long long #define repne(x,y,i) for(i=(x);i<(y);++i) #define repe(x,y,i) for(i=(x);i<=(y);++i) #define repde(x,y,i) for(i=(x);i>=(y);--i) #define repdne(x,y,i) for(i=(x);i>(y);--i) #define ri register int inline void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr=='-')sign=-1; for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;} inline void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if (chr=='-')sign=-1; for(n=0;dight(chr);chr=getchar())n=n*10+chr-'0';n*=sign;} int mu[105],prim[105],len; ll n; bool isprim[105]; void init(){ for(int i = 1;i <= 100;++i) isprim[i] = true; for(int i = 2;i <= 100;++i){ if(isprim[i]) prim[len++] = i,mu[i] = -1; for(int j = 0;j < len && i * prim[j] <= 100;++j){ mu[i*prim[j]] = -mu[i]; isprim[i*prim[j]] = false; if(i % prim[j] == 0){ mu[i*prim[j]] = 0; break; } } } } int main() { freopen("a.in","r",stdin); freopen("b.out","w",stdout); init(); T{ Read(n); ll sum = 0; for(int i = 2;i <= 60;++i){ sum = sum + (ll)mu[i] * ((ll)pow((long double)n+0.1,(long double)1.0/i)-1);//减掉等于1的情况 } printf("%lld\n",sum+n-1); } return 0; }