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 = ;
 bool check[N+];
 int prime[N+];
 int mu[N+];
 ll mu_sum[N+];
 void Moblus(){
     CLR(check, false);
     mu[] = ;
     int tot = ;
     for(int i = ; i <= N; i++){
         if( !check[i] ){
             prime[tot++] = i;
             mu[i] = -;
         }
         for(int j = ; j < tot && i*prime[j] <= N; j++){
             check[i * prime[j]] = true;
             if( i % prime[j] == ){
                 mu[i * prime[j]] = ;
                 break;
             }
             else
                 mu[i * prime[j]] = -mu[i];
         }
     }
     for(int i = ; i <= N; ++i)
         mu_sum[i] = mu_sum[i-] + 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 = , ed;
     for(ll i = ; i <= n; i = ed+){
         ed = n/(n/i);
         ans -= (ed - i + )*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-));
     return ;
 }