HRY and mobius
对于k = 0 ,显然最后答案为,n。
对于k = 1,答案就是
这就是经典的杜教筛问题了。
对于k = 2
/*
Author : lifehappy
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e7 + 10;
int prime[N], mu[N], sum[N], cnt;
bool st[N];
void init() {
mu[1] = 1;
for(int i = 2; i < N; i++) {
if(!st[i]) {
prime[++cnt] = i;
mu[i] = -1;
}
for(int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {
st[i * prime[j]] = 1;
if(i % prime[j] == 0) {
break;
}
mu[i * prime[j]] = -mu[i];
}
}
for(int i = 1; i < N; i++) {
sum[i] = mu[i] + sum[i - 1];
}
}
unordered_map<ll, ll> ans_s;
ll Djs(ll n) {
if(n < N) return sum[n];
if(ans_s.count(n)) return ans_s[n];
ll ans = 1;
for(ll l = 2, r; l <= n; l = r + 1) {
r = n / (n / l);
ans -= (r - l + 1) * Djs(n / l);
}
return ans_s[n] = ans;
}
ll calc(ll n) {
ll ans = 0;
for(ll i = 1; i * i <= n; i++) {
ans += mu[i] * (n / i / i);
}
return ans;
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
init();
int T;
scanf("%d", &T);
while(T--) {
ll n, k;
scanf("%lld %lld", &n, &k);
if(k == 0) {
printf("%lld\n", n);
}
else if(k & 1) {
printf("%lld\n", Djs(n));
}
else {
printf("%lld\n", calc(n));
}
}
return 0;
} 
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