题目链接：VJ(private)
Euler Sieve
题目描述
定义函数f为约数和函数和欧拉函数的卷积。求f前n项和。
1 <= n <= 5e6
解释
用欧拉筛（线性筛）筛素数的同时求出每项的欧拉函数和约数和函数。用双重循环将f的前n项和直接累加。
代码部分
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| #include<bits/stdc++.h>
using namespace std;
int prime[5050000];
bool flag[5050000];
int phi[5050000];
int f[5050000];
int g[5050000];
int cnt;
int main()
{
int n;
cin>>n;
flag[1]=true;
phi[1]=f[1]=g[1]=1;
for (int i=1;i<=n;i++) {
if (!flag[i]) {
prime[++cnt]=i;
phi[i]=i-1;
f[i]=i+1;
g[i]=i+1;
}
for (int j=1;j<=cnt&&i*prime[j]<=n;j++) {
flag[i*prime[j]]=true;
if (i%prime[j]==0) {
phi[i*prime[j]]=phi[i]*prime[j];
g[i*prime[j]]=g[i]*prime[j]+1;
f[i*prime[j]]=f[i]/g[i]*g[i*prime[j]];
break;
}
phi[i*prime[j]]=phi[i]*(prime[j]-1);
f[i*prime[j]]=f[i]*f[prime[j]];
g[i*prime[j]]=1+prime[j];
}
}
long long ans=0;
for (int i=1;i<=n;i++)
for (int j=1;j<=n/i;j++)
ans+=phi[i]*f[j];
cout<<ans<<endl;
return 0;
}
|
补充
积性函数可以用欧拉筛来求出每一项值。常见的积性函数有欧拉函数、莫比乌斯函数、约数个数函数、约数和函数。
以下是oi-wiki中求四种积性函数的代码。
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|
void pre() {
memset(is_prime, 1, sizeof(is_prime));
int cnt = 0;
is_prime[1] = 0;
phi[1] = 1;
for (int i = 2; i <= 5000000; i++) {
if (is_prime[i]) {
prime[++cnt] = i;
phi[i] = i - 1;
}
for (int j = 1; j <= cnt && i * prime[j] <= 5000000; j++) {
is_prime[i * prime[j]] = 0;
if (i % prime[j])
phi[i * prime[j]] = phi[i] * phi[prime[j]];
else {
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
}
}
}
|
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void pre() {
mu[1] = 1;
for (int i = 2; i <= 1e7; ++i) {
if (!v[i]) mu[i] = -1, p[++tot] = i;
for (int j = 1; j <= tot && i <= 1e7 / p[j]; ++j) {
v[i * p[j]] = 1;
if (i % p[j] == 0) {
mu[i * p[j]] = 0;
break;
} else {
mu[i * p[j]] = -mu[i];
}
}
}
}
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void pre() {
d[1] = 1;
for (int i = 2; i <= n; ++i) {
if (!v[i]) v[i] = 1, p[++tot] = i, d[i] = 2, num[i] = 1;
for (int j = 1; j <= tot && i <= n / p[j]; ++j) {
v[p[j] * i] = 1;
if (i % p[j] == 0) {
num[i * p[j]] = num[i] + 1;
d[i * p[j]] = d[i] / num[i * p[j]] * (num[i * p[j]] + 1);
break;
} else {
num[i * p[j]] = 1;
d[i * p[j]] = d[i] * 2;
}
}
}
}
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void pre() {
g[1] = f[1] = 1;
for (int i = 2; i <= n; ++i) {
if (!v[i]) v[i] = 1, p[++tot] = i, g[i] = i + 1, f[i] = i + 1;
for (int j = 1; j <= tot && i <= n / p[j]; ++j) {
v[p[j] * i] = 1;
if (i % p[j] == 0) {
g[i * p[j]] = g[i] * p[j] + 1;
f[i * p[j]] = f[i] / g[i] * g[i * p[j]];
break;
} else {
f[i * p[j]] = f[i] * f[p[j]];
g[i * p[j]] = 1 + p[j];
}
}
}
}
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