[BZOJ3944]Sum（杜教筛）

题目描述

传送门

题解

杜教筛裸题
我不会手写hash表…
讲解移步：http://blog.csdn.net/clove_unique/article/details/66991109

代码

#include
#include
#include
#include
#include
#include
using namespace std;
#define N 5000000
#define LL long long

int T,n;
int p[N+3],prime[N+3];LL phi[N+3],mu[N+3];
map <int,LL> ansphi,ansmu;

void get()
{
    phi[1]=mu[1]=1;
    for (int i=2;i<=N;++i)
    {
        if (!p[i])
        {
            prime[++prime[0]]=i;
            phi[i]=i-1;
            mu[i]=-1;
        }
        for (int j=1;j<=N&&i*prime[j]<=N;++j)
        {
            p[i*prime[j]]=1;
            if (i%prime[j]==0)
            {
                phi[i*prime[j]]=phi[i]*prime[j];
                mu[i*prime[j]]=0;
                break;
            }
            else
            {
                phi[i*prime[j]]=phi[i]*phi[prime[j]];
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
    for (int i=1;i<=N;++i) phi[i]+=phi[i-1],mu[i]+=mu[i-1];
}
LL sumphi(LL n)
{
    if (nreturn phi[n];
    if (ansphi[n]) return ansphi[n];
    LL tmp;
    if (n&1) tmp=(n+1)/2*n;
    else tmp=n/2*(n+1);
    for (int i=2,j=0;i<=n;i=j+1)
    {
        j=n/(n/i);
        tmp-=sumphi(n/i)*(LL)(j-i+1);
        if (j==n) break;
    }
    ansphi[n]=tmp;
    return tmp;
}
LL summu(int n)
{
    if (n<=N) return mu[n];
    if (ansmu[n]) return ansmu[n];
    LL tmp=1;
    for (int i=2,j=0;i<=n;i=j+1)
    {
        j=n/(n/i);
        tmp-=summu(n/i)*(LL)(j-i+1);
        if (j==n) break;
    }
    ansmu[n]=tmp;
    return tmp;
}
int main()
{
    get();
    scanf("%d",&T);
    while (T--)
    {
        scanf("%lld",&n);
        printf("%lld %lld\n",sumphi(n),summu(n));
    }
}