BZOJ 2818(莫比乌斯反演)

最近刚看莫比乌斯反演  数学不好是硬伤啊 不过总算跑出来了

第一种是没优化的跑这题

#include 
#include 
using namespace std;
const int maxn = 1e7 + 5;
int isprime[maxn];
int prime[maxn];
int mu[maxn];
int cnt;
void mobi(int n)
{
    memset(isprime, 0, sizeof isprime);
    mu[1] = 1;
    cnt = 0;
    for(int i = 2; i <= n; i++)
    {
        if(!isprime[i]){
            prime[cnt++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < cnt && i * prime[j] <= n; j++)
        {
            isprime[i * prime[j]] = 1;
            if(i % prime[j]) mu[i * prime[j]] = -mu[i];
            else{
                mu[i * prime[j]] = 0;
                break;
            }
        }
    }
}
int main()
{
    int a;
    long long ans = 0;
    cin>>a;
    mobi(a);
    for(int j = 0; j < cnt && prime[j] <= a; j++)
    {
        long long tmp = a / prime[j];
        for(int i = 1; i <= tmp; i++)
            ans += 1ll * mu[i] * ((tmp/i) * (tmp/i));
    }
    cout<

第二种是优化过

#include 
#include 
using namespace std;
const int maxn = 1e7 + 5;
int isprime[maxn];
int prime[maxn];
int g[maxn];
int sum[maxn];
int mu[maxn];
int cnt;
void mobi(int n)
{
    memset(isprime, 0, sizeof isprime);
    memset(g, 0, sizeof g);
    mu[1] = 1;
    cnt = 0;
    for(int i = 2; i <= n; i++)
    {
        if(!isprime[i]){
            prime[cnt++] = i;
            mu[i] = -1;
            g[i] = 1;
        }
        for(int j = 0; j < cnt && i * prime[j] <= n; j++)
        {
            isprime[i * prime[j]] = 1;
            if(i % prime[j])
            {
                mu[i * prime[j]] = -mu[i];
                g[i * prime[j]] = mu[i] - g[i];
            }
            else{
                mu[i * prime[j]] = 0;
                g[i * prime[j]] = mu[i];
                break;
            }
        }
    }
    sum[0] = 0;
    for(int i = 1; i <= n; i++)
        sum[i] = sum[i - 1] + g[i];
    
}
int main()
{
    int a;
    long long ans = 0;
    cin>>a;
    mobi(a);
    for(int i = 1, last; i <= a; i = last + 1)
    {
        last = a/(a/i);
        ans += (long long)(sum[last] - sum[i - 1]) * (a/i) * (a/i);
    }
    cout<