51nod 1244 莫比乌斯函数之和(积性函数前缀和)

关于积性函数前缀和的问题，可以关注糖老师的博客
http://blog.csdn.net/skywalkert/article/details/50500009
推导不写了
结论是M(n)=∑ni=1u(i)
M(n)=1−∑ni=2M(ni)
直接递归求解，对于n，直接sqrt(n)的复杂度枚举i，然后递归求解
蓝儿还是会T，用map记忆化一下，小于500W的前缀和可以直接预处理，能过

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代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           5000005
#define   MAXN          1000005
#define   maxnode       105
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
const double pi    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-8;
const LL     mod   = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
 }
/*****************************************************/

bool prime[MAX];
int pr[MAX];
int tot;
int mu[MAX];
map ma;
void init(){
    mem(prime,0);tot=0;
    mu[0]=0;mu[1]=1;
    for(int i=2;iif(!prime[i]){
            pr[tot++]=i;
            mu[i]=-1;
        }
        for(int j=0;j1;
            if(i%pr[j]==0){
                mu[i*pr[j]]=0;
                break;
            }
            mu[i*pr[j]]=-mu[i];
        }
    }
    for(int i=1;i1];
}
LL cal(LL a){
    if(areturn mu[a];
    if(ma.count(a)) return ma[a];
    LL ans=1;
    for(LL i=2;i<=a;){
        LL x=a/(a/i);
        ans-=cal(a/i)*(x-i+1);
        i=x+1;
    }
    ma[a]=ans;
    return ans;
}
int main(){
    //freopen("in.txt","r",stdin);
    LL a,b;
    init();
    ma.clear();
    cin>>a>>b;
    cout<1)<return 0;
}