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@ -1,21 +1,30 @@
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#include <bits/stdc++.h>
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using namespace std;
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#define LL long long
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const int N = 100 + 100;
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#define int long long
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#define endl "\n"
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const int N = 110;
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/*
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10
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36
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1000000000000000000
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4
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9
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1001003332
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*/
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// 筛法求莫比乌斯函数
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LL mu[N], sum[N];
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int mu[N], sum[N];
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int primes[N], cnt;
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bool st[N];
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void get_mobius(LL n) {
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void get_mobius(int n) {
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mu[1] = 1;
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for (LL i = 2; i <= n; i++) {
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for (int i = 2; i <= n; i++) {
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if (!st[i]) {
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primes[cnt++] = i;
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mu[i] = -1;
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}
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for (LL j = 0; primes[j] <= n / i; j++) {
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LL t = primes[j] * i;
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for (int j = 0; primes[j] <= n / i; j++) {
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int t = primes[j] * i;
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st[t] = true;
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if (i % primes[j] == 0) {
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mu[t] = 0;
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@ -26,18 +35,18 @@ void get_mobius(LL n) {
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}
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// 维护u(x)前缀和
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// 这玩意有啥用?
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for (LL i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
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for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
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}
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int main() {
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signed main() {
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// 筛法求莫比乌斯函数
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get_mobius(N - 1);
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LL T;
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int T;
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while (cin >> T) {
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LL s = 1;
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for (LL i = 2; i <= 64; i++) s -= mu[i] * (LL)(pow(T * 1.0, 1.0 / i) - 1);
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int s = 1;
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for (int i = 2; i <= 64; i++) s -= mu[i] * (int)(pow(T * 1.0, 1.0 / i) - 1);
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cout << s << endl;
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}
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return 0;
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}
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