SPOJ DIVCNT2

来源：互联网发布：什么是噪声数据编辑：程序博客网时间：2024/06/09 14:28

题目链接：
http://www.spoj.com/problems/DIVCNT2/

根据rzz的课件可以分三段做

度教搞出来一种分一次做的方法看起来很神的样子

#include<cstdio>#include<iostream>#include<vector>#include<cstring>#include<cstdlib>#include<algorithm>#include<map>using namespace std;#define ll long longchar c;inline void read(ll&a){a=0;do c=getchar();while(c<'0'||c>'9');while(c<='9'&&c>='0')a=(a<<3)+(a<<1)+c-'0',c=getchar();}ll n;const    ll MAXN=(int)1e8+1;map<ll,ll >Mu,W,O;int mu[MAXN],tot,prime[5761490],w[MAXN];int cnt;int D[2049];inline ll m(ll x){    return mu[x]==mu[x-1]?0:(D[w[x]-w[x-1]]&1?-1:1);}ll SolveMu(ll n){    if(n<MAXN)return mu[n];    if(Mu[n])return Mu[n];    ll res=0;    for(ll i=1;i*1ll*i<=n;i++)        (res+=m(i)*(n/(i*i)));    return Mu[n]=res;}ll SolveW(ll n){    if(n<MAXN)return w[n];    if(W[n])return W[n];    ll res=0;    ll i=1,j;    while(i<=n)    {        j=n/(n/i);        (res+=(SolveMu(j)-SolveMu(i-1))*1ll*((n/i)));        i=j+1;    }    return W[n]=res;}ll SolveO(ll n){    if(O[n])return O[n];    ll res=0;    ll i=1,j;    while(i<=n)    {        j=n/(n/i);        (res+=(SolveW(j)-SolveW(i-1))*1ll*((n/i)));        i=j+1;    }    return O[n]=res;}void out(ll x){    if(x<0)putchar('-'),x=-x;    if(x>9)out(x/10);    putchar('0'+x%10);}ll Up;vector<ll>Q;int main(){    Up=-1;    ll T;    mu[1]=w[1]=1;    read(T);    while(T--)    {        ll n;        read(n);        Q.push_back(n);        if(Up<n)Up=n;        //out(SolveO(n));        //puts("");    }       Up=min(Up+1,MAXN);    for(int i=2;i<Up;i++)    {        if(!w[i])prime[++tot]=i,mu[i]=-1,w[i]=2;        for(int j=1,k;j<=tot&&(k=prime[j]*i)<Up;j++)        {            if(i%prime[j]==0)            {mu[k]=0;w[k]=w[i];break;}            mu[k]=-mu[i];w[k]=w[i]*2;        }        mu[i]=mu[i-1]+mu[i]*mu[i];        w[i]+=w[i-1];    }    int j=1;    for(int i=0;i<=11;j*=2,i++)D[j]=i;    for(int i=0;i<Q.size();i++)        out(SolveO(Q[i])),puts("");    return 0;}

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