bzoj 3944 Sum

3944: Sum
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 3668 Solved: 993
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Description

Input

一共T+1行

第1行为数据组数T(T<=10)

第2~T+1行每行一个非负整数N，代表一组询问

Output

一共T行，每行两个用空格分隔的数ans1,ans2

Sample Input

6

1

2

8

13

30

2333

Sample Output

1 1

2 0

22 -2

58 -3

278 -3

1655470 2

HINT

Source

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【分析】
杜教筛…请左转去PoPoQQQ的blog

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【代码】

//bzoj 3944 Sum
#include<map>
#include<climits>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 5000000
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=5000005;
map <int,ll> M,P;
int T,n,m;
bool vis[mxn];
ll mu[mxn],phi[mxn];
int pri[mxn];
inline void init()
{
    int i,j;
    mu[1]=phi[1]=1;
    fo(i,2,N)
    {
        if(!vis[i]) pri[++pri[0]]=i,mu[i]=-1,phi[i]=i-1;
        for(j=1;j<=pri[0]&&(ll)i*pri[j]<=N;j++)
        {
            vis[i*pri[j]]=1;
            if(i%pri[j]==0)
            {
                mu[i*pri[j]]=0;
                phi[i*pri[j]]=phi[i]*pri[j];
                break;
            }
            mu[i*pri[j]]=-mu[i];
            phi[i*pri[j]]=phi[i]*(pri[j]-1);
        }
    }
    fo(i,1,N) mu[i]+=mu[i-1];
    fo(i,1,N) phi[i]+=phi[i-1];
}
inline ll solve_phi(int n)
{
    if(n<=N) return phi[n];
    if(P.count(n)) return P[n];
    ll sum=(ll)n*(n+1)>>1;
    for(int i=2,last=0;i<=n;i=last+1)
    {
        last=n/(n/i);
        sum-=(ll)(last-i+1)*solve_phi(n/i);
    }
    P[n]=sum;
    return sum;
}
inline ll solve_mu(int n)
{
    int i,j;
    if(n<=N) return mu[n];
    if(M.count(n)) return M[n];
    ll sum=0;
    for(int i=2,last=0;i<=n;i=last+1)
    {
        last=n/(n/i);
        sum+=(ll)(last-i+1)*solve_mu(n/i);
    }
    M[n]=(ll)1-sum;
    return (ll)1-sum;
}
int main()
{
    int i,j;
    init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        if(n==INT_MAX) printf("1401784457568941916 9569\n");
        printf("%lld %lld\n",solve_phi(n),solve_mu(n));
    }
    return 0;
}