SPOJ 20713 DIVCNT2 - Counting Divisors (square)

DIVCNT2 - Counting Divisors (square)

#sub-linear #dirichlet-generating-function

Let \sigma_0(n)σ0(n) be the number of positive divisors of nn.

For example, \sigma_0(1) = 1σ0(1)=1, \sigma_0(2) = 2σ0(2)=2 and \sigma_0(6) = 4σ0(6)=4.

LetS_2(n) = \sum _{i=1}^n \sigma_0(i^2).S2(n)=i=1∑nσ0(i2).

Given NN, find S_2(N)S2(N).

Input

First line contains TT (1 \le T \le 100001≤T≤10000), the number of test cases.

Each of the next TT lines contains a single integer NN. (1 \le N \le 10^{12}1≤N≤1012)

Output

For each number NN, output a single line containing S_2(N)S2(N).

Example

Input

512310100

Output

147481194

Explanation for Input

- S_2(3) = \sigma_0(1^2) + \sigma_0(2^2) + \sigma_0(3^2) = 1 + 3 + 3 = 7S2(3)=σ0(12)+σ0(22)+σ0(32)=1+3+3=7

Information

There are 6 Input files.

- Input #1: 1 \le N \le 100001≤N≤10000, TL = 1s.

- Input #2: 1 \le T \le 800,\ 1 \le N \le 10^{8}1≤T≤800, 1≤N≤108, TL = 20s.

- Input #3: 1 \le T \le 200,\ 1 \le N \le 10^{9}1≤T≤200, 1≤N≤109, TL = 20s.

- Input #4: 1 \le T \le 40,\ 1 \le N \le 10^{10}1≤T≤40, 1≤N≤1010, TL = 20s.

- Input #5: 1 \le T \le 10,\ 1 \le N \le 10^{11}1≤T≤10, 1≤N≤1011, TL = 20s.

- Input #6: T = 1,\ 1 \le N \le 10^{12}T=1, 1≤N≤1012, TL = 20s.

My C++ solution runs in 5.3 sec. (total time)

Source Limit is 6 KB.

很迷的函数题。

如何求 d(i^2)?

d(i^2)= (2*a1+1)(2*a2+1)(2*a3+1)...(2*ak+1)

我们考虑一下选哪些质因子的集合，上式

=Σ2^|S| *π a[i] ,i属于S

=Σ(p|i) 2^w(p)。

其中w(x)为x的质因子数。

然后发现2^w(x)=Σ(i|x) μ^2(i)

所以ANS= Σμ^2(i) *Σd(j) ,其中1<=i<=n,1<=j<=(n/i)。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int zs[10000005],t=0,T,sq[50000005];
int miu[50000005],low[50000005],maxn;
bool v[50000005];
ll d[50000005],n;inline void init(){miu[1]=1,d[1]=1,low[1]=1;for(int i=2;i<=maxn;i++){if(!v[i]) zs[++t]=i,miu[i]=-1,d[i]=2,low[i]=i;for(int j=1,u;j<=t&&(u=zs[j]*i)<=maxn;j++){v[u]=1;if(!(i%zs[j])){low[u]=low[i]*zs[j];if(low[i]==i) d[u]=d[i]+1;else d[u]=d[low[u]]*d[i/low[i]];break;}low[u]=zs[j];d[u]=d[i]<<1;miu[u]=-miu[i];}}for(int i=1;i<=maxn;i++) d[i]+=d[i-1];for(int i=1;i<=maxn;i++) sq[i]=sq[i-1]+miu[i]*miu[i];
}inline ll getsq(ll x){if(x<=maxn) return sq[x];ll an=0;for(int i=1;i*(ll)i<=x;i++){an+=miu[i]*(x/(i*(ll)i));}return an;
}inline ll getd(ll x){if(x<=maxn) return d[x];ll an=0;for(ll i=1,j,now;i<=x;i=j+1){now=x/i,j=x/now;an+=(j-i+1)*now;}return an;
}inline ll query(ll x){ll an=0;for(ll i=1,j,now;i<=x;i=j+1){now=x/i,j=x/now;an+=(getsq(j)-getsq(i-1))*getd(now);}return an;
}int main(){scanf("%d",&T);if(T>800) maxn=1000000;else maxn=50000000;init();while(T--){scanf("%lld",&n);printf("%lld\n",query(n));}return 0;
}

转载于:https://www.cnblogs.com/JYYHH/p/8511216.html