20160322 HDU5317 RGCDQ（筛数+树状数组）

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题意：

F(n)是n的不同素因子的个数，求在[ L，R ]，区间内最大的gcd(F(i),F(j))。

思路：

看上去是数论的数据结构题......

可以注意得到数据范围小于1E6，意味着F(n)的取值范围很小（不会超过7）。

很容易想到记每个位置的F(n)，在查询区间时返回区间内各个F(n)的个数，然后判断gcd值的方法。

用素数表将范围内所有的F(n)筛选出来，通过树状数组来求区间和。

#include <iostream>#include <stdio.h>#include <math.h>using namespace std;int pri[168]= {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997};int a[1000000+5],k[1000000+5];int c[8][1000000+5];inline int lowbit(int x){    return (x&-x);}void add(int x,int d){    for(int i=x; i<=1000005; i+=lowbit(i))        c[d][i]+=1;}int get(int x,int d){    int sum=0;    for(int i=x; i; i-=lowbit(i))        sum+=c[d][i];    return sum;}void solve(){    for(int i=0; i<1000005; i++)        k[i]=i;    for(int i=0; i<168; i++)    {        int c=pri[i];        while(c<1000005)        {            a[c]++;            while(k[c]%pri[i]==0)            {                k[c]/=pri[i];            }            c+=pri[i];        }    }    for(int i=2; i<1000005; i++)    {        if(k[i]!=1)            a[i]++;        add(i,a[i]);    }}int main(){    solve();    //cout<<a[570570]<<endl;    int T,l,r;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&l,&r);        int d[17];        for(int i=1; i<=7; i++)            d[i]=get(r,i)-get(l-1,i);        /*int sum=0;        for(int i=0;i<8;i++)        {            cout<<"test "<<i<<endl;            cout<<get(r,i)<<" "<<get(l-1,i)<<endl;            cout<<d[i]<<endl;            sum+=d[i];        }        cout<<sum<<endl;*/        if(d[7]>=2)        {            printf("7\n");        }        else if(d[6]>=2)        {            printf("6\n");        }        else if(d[5]>=2)        {            printf("5\n");        }        else if(d[4]>=2)        {            printf("4\n");        }        else if(d[3]>=2||(d[6]==1&&d[3]==1))        {            printf("3\n");        }        else if(d[2]>=2||(d[6]==1&&d[4]==1)||(d[6]==1&&d[2]==1)||(d[4]==1&&d[2]==1))        {            printf("2\n");        }        else        {            printf("1\n");        }    }    return 0;}

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