本文借鉴了该博客莫比乌斯函数和
这篇博客里面求解的是一个区间段内的莫比乌斯函数和,稍微改一下就可以计算
#include<bits/stdc++.h>
const int INF = 0x3f3f3f3f;
const int Maxn = 1e7 * 2;
const int mod = 2333333;
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
int pcnt = 0, prime[1300000]; // 质数
short mu[Maxn]; // 莫比乌斯函数值
bool vis[Maxn];
void init() {
mu[1] = 1;
for (int i = 2; i < Maxn; i++) {
if (vis[i] == 0) {
mu[i] = -1;
prime[++pcnt] = i;
}
for (int j = 1; j <= pcnt && i * prime[j] < Maxn; j++) {
vis[i * prime[j]] = 1;
if (i % prime[j] != 0)
mu[i * prime[j]] = -mu[i];
else {
mu[i * prime[j]] = 0;
break;
}
}
}
for (int i = 2; i < Maxn; i++) mu[i] += mu[i - 1]; // 函数前缀和
}
struct Hash {
long long key;
int value, next;
} node[mod];
int cnt = 0, Head[mod] = {
0};
void Insert(long long N, int v) {
// 记忆
int ha = N % mod;
node[++cnt].key = N;
node[cnt].value = v;
node[cnt].next = Head[ha];
Head[ha] = cnt;
}
int M(long long N) {
if (N < Maxn) return mu[N];
int ha = N % mod;
for (int i = Head[ha]; i != 0; i = node[i].next) {
if (node[i].key == N) // 如果已经计算过
return node[i].value;
}
int ans = 0;
for (long long i = 2, j; i <= N; i = j + 1) {
j = N / (N / i); // 分块加速
ans += (j - i + 1) * M(N / i);
}
Insert(N, 1 - ans); // 做记忆
return 1 - ans;
}
int main() {
init();
long long n;
while (cin >> n) {
cout << M(n) << endl;
}
}