A Simple Math Problem
题目大意
输入描述
输出描述
样例
输入
3
输出
5
推导
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+1;
int prime[maxn], mu[maxn], phi[maxn], tot;
bool vis[maxn];
int f[maxn];
void get_pmp(int n) {
vis[1] = mu[1] = phi[1] = 1;
for(int i = 2; i <= n; ++ i) {
if (!vis[i]) {
prime[++tot] = i;
mu[i] = -1;
phi[i] = i-1;
}
for(int j = 1; prime[j] <= n/i; ++ j) {
vis[prime[j]*i] = 1;
phi[prime[j]*i] = prime[j] * phi[i];
if (i % prime[j] == 0) break;
mu[prime[j]*i] = mu[prime[j]] * mu[i];
phi[prime[j]*i] = phi[prime[j]] * phi[i];
}
}
}
void get_f(int n) {
for(int i = 1; i <= n; ++ i) {
int t = i;
while(t) {
f[i] += t%10;
t /= 10;
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
get_pmp(n);
get_f(n);
long long res = 1;
for(int i = 1; i <= n; ++ i) {
long long t = 0;
for(int j = 1; j <= n/i; ++ j) {
t += f[j*i];
}
res += mu[i] * n/i * t - phi[i] * f[i];
}
cout << res << "
";
return 0;
}
代码