《洛谷P4213 【模板】杜教筛（Sum）》

杜教筛的模板题：

推出式子就可以了。

具体的怎么推放到杜教筛里讲。

这题主要常数卡得厉害，把预处理的范围开大后冲过去了

还有两种情况，n * (n + 1) 可能会爆long long。

然后r = 2147483647时，L + 1就爆int了，所以也要特判。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
typedef unsigned long long ULL;
const int N = 5e6+5;
const int M = 1e6+5;
const LL Mod = 9999999967;
#define rg register
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read()
    {
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

int m = 5000000;
int prime[N],tot = 0;
LL phi[N],mu[N];
bool vis[N];
unordered_map<int,LL> mp1,mp2;
void init()
{
    mu[1] = 1,phi[1] = 1;
    for(rg int i = 2;i <= m;++i)
    {
        if(!vis[i])
        {
            prime[++tot] = i;
            phi[i] = i - 1;
            mu[i] = -1;
        }
        for(rg int j = 1;j <= tot && prime[j] * i <= m;++j)
        {
            vis[prime[j] * i] = 1;
            if(i % prime[j] == 0){phi[prime[j] * i] = phi[i] * prime[j];break;}
            else 
            {
                mu[prime[j] * i] = -mu[i];
                phi[prime[j] * i] = phi[prime[j]] * phi[i];
            }
        }
    }
    for(rg int i = 1;i <= m;++i) phi[i] += phi[i - 1];
    for(rg int i = 1;i <= m;++i) mu[i] += mu[i - 1];
}
LL solve1(int n)//欧拉函数前缀和
{
    if(n <= m) return phi[n];
    if(mp1[n]) return mp1[n];
    LL ans = 0;
    for(rg int L = 2,r = 0;L <= n && r < 2147483647;L = r + 1)
    {
        r = n / (n / L);
        ans += (r - L + 1) * solve1(n / L);
    }
    return mp1[n] = ((ULL)n * (n + 1) / 2) - ans;
}
LL solve2(int n)//莫比乌斯函数前缀和
{
    if(n <= m) return mu[n];
    if(mp2[n]) return mp2[n];
    LL ans = 1;
    for(rg int L = 2,r = 0;L <= n && r < 2147483647;L = r + 1)
    {
        r = n / (n / L);
        ans -= (r - L + 1) * solve2(n / L);
    }
    return mp2[n] = ans;
}
int main()
{
    init();
    int ca;ca = read();
    while(ca--)
    {
        int n;n = read();
        printf("%lld %lld
",solve1(n),solve2(n));
    }
    system("pause");
    return 0;
}