Relatively Prime Powers

CodeForces - 1036F

Consider some positive integer xx. Its prime factorization will be of form x=2k1?3k2?5k3?…x=2k1?3k2?5k3?…

Let's call xx elegant if the greatest common divisor of the sequence k1,k2,…k1,k2,… is equal to 11. For example, numbers 5=515=51, 12=22?312=22?3, 72=23?3272=23?32 are elegant and numbers 8=238=23 (GCD=3GCD=3), 2500=22?542500=22?54 (GCD=2GCD=2) are not.

Count the number of elegant integers from 22 to nn.

Each testcase contains several values of nn, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer TT (1≤T≤1051≤T≤105) — the number of values of nn in the testcase.

Each of the next TT lines contains a single integer nini (2≤ni≤10182≤ni≤1018).

Output

Print TT lines — the ii-th line should contain the number of elegant numbers from 22to nini.

Example

Input

4427210

Output

21616

Note

Here is the list of non-elegant numbers up to 1010:

• 4=22,GCD=24=22,GCD=2;
• 8=23,GCD=38=23,GCD=3;
• 9=32,GCD=29=32,GCD=2.

The rest have GCD=1GCD=1.

題意：

給你一個大于等于2的整數N

讓你求2~N 中有多少個整數x，

唯一分解后質因子的冪次分別是e1,e2,e3, 時 gcd(e1,e2,e3)=1

思路：

正難則反，一共有N-1個數，我們只需要減去那些gcd不為1的即可，

我們可以分別枚舉gcd為2,3,4,5.,,,, 等等

根據容斥原理，gcd 為i時，他對答案的貢獻即為 mu[i]*(n^(1/i) -1 ) mu是莫比烏斯函數。

至于系數為什么恰好是莫比烏斯函數，可以先學這篇博客感受一下：

https://www.cnblogs.com/qieqiemin/p/11537681.html

那么我們來看n^(1/i) -1 是2~n中，質因子分解冪次都為i的數的個數。

即n開i次方-1，先去的1就是就是一個數開任何次方都>=1,數字1被算進去了，需要減去。

細節見代碼：

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define sz(a) int(a.size()) #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl #define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c)) #define du2(a,b) scanf("%d %d",&(a),&(b)) #define du1(a) scanf("%d",&(a)); using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;} void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}} void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}inline void getInt(int *p); const int maxn = 1000010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ long long gen(long long n, long long k) {long long t = powl(n, 1. / k) - 0.5;return t + (powl(t + 1, k) - 0.5 <= n); } #define N maxn bool vis[N]; long long prim[N], mu[N], sum[N], cnt; void get_mu(long long n) {mu[1] = 1;for (long long i = 2; i <= n; i++) {if (!vis[i]) {mu[i] = -1; prim[++cnt] = i;}for (long long j = 1; j <= cnt && i * prim[j] <= n; j++) {vis[i * prim[j]] = 1;if (i % prim[j] == 0) { break; }else { mu[i * prim[j]] = -mu[i]; }}}for (long long i = 1; i <= n; i++) { sum[i] = sum[i - 1] + mu[i]; } } int main() {//freopen("D:\\code\\text\\input.txt","r",stdin);//freopen("D:\\code\\text\\output.txt","w",stdout);int t;get_mu(maxn - 1);du1(t);while (t--) {ll n;scanf("%lld", &n);ll ans = n - 1;for (ll i = 2ll; i <= 64ll; ++i) {ans += mu[i] * (gen(n, i) - 1ll);}printf("%lld\n", ans );}return 0; }inline void getInt(int *p) {char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');if (ch == '-') {*p = -(getchar() - '0');while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 - ch + '0';}} else {*p = ch - '0';while ((ch = getchar()) >= '0' && ch <= '9') {*p = *p * 10 + ch - '0';}} }

轉載于:https://www.cnblogs.com/qieqiemin/p/11543610.html

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