51nod 1244 莫比乌斯函数之和

题目链接：51nod 1244 莫比乌斯函数之和

题解参考syh学长的博客：http://www.cnblogs.com/AOQNRMGYXLMV/p/4932537.html %%%

关于这一类求积性函数前缀和的方法，学习参考博客：http://blog.csdn.net/skywalkert/article/details/50500009 

要好好看大神的博客哦orz

用筛法预处理前N^(2/3)项，后面的记忆化搜索解决。

不太会用哈希，就用map记忆化一下：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
using namespace std;
#define CLR(a,b) memset((a),(b),sizeof((a)))
typedef long long ll;
const int N = 5000000;
bool check[N+10];
int prime[N+10];
int mu[N+10];
ll mu_sum[N+10];
void Moblus(){
    CLR(check, false);
    mu[1] = 1;
    int tot = 0;
    for(int i = 2; i <= N; i++){
        if( !check[i] ){
            prime[tot++] = i;
            mu[i] = -1;
        }
        for(int j = 0; j < tot && i*prime[j] <= N; j++){
            check[i * prime[j]] = true;
            if( i % prime[j] == 0){
                mu[i * prime[j]] = 0;
                break;
            }
            else
                mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1; i <= N; ++i)
        mu_sum[i] = mu_sum[i-1] + mu[i];
}
map<ll,ll> mp;
ll Mertens(ll n){
    if(n <= N) return mu_sum[n];
    if(mp.count(n)) return mp[n];
    ll ans = 1, ed;
    for(ll i = 2; i <= n; i = ed+1){
        ed = n/(n/i);
        ans -= (ed - i + 1)*Mertens(n/i);
    }
    return mp[n] = ans;
}
int main(){
    Moblus();
    ll a, b;
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", Mertens(b) - Mertens(a-1));
    return 0;
}