HDU 5317(RGCDQ-统计)

RGCDQ

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1963    Accepted Submission(s): 830

Problem Description

Mr. Hdu is interested in Greatest Common Divisor (GCD). He wants to find more and more interesting things about GCD. Today He comes up with Range Greatest Common Divisor Query (RGCDQ). What’s RGCDQ? Please let me explain it to you gradually. For a positive integer x, F(x) indicates the number of kind of prime factor of x. For example F(2)=1. F(10)=2, because 10=2*5. F(12)=2, because 12=2*2*3, there are two kinds of prime factor. For each query, we will get an interval [L, R], Hdu wants to know maxGCD(F(i),F(j)) (L≤i<j≤R)

 

Input

There are multiple queries. In the first line of the input file there is an integer T indicates the number of queries.
In the next T lines, each line contains L, R which is mentioned above.

All input items are integers.
1<= T <= 1000000
2<=L < R<=1000000

 

Output

For each query，output the answer in a single line.
See the sample for more details.

 

Sample Input

   
   
   
   

    
    
    
    2
2 3
3 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
1
   
   
   
   

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:   5405  5404  5403  5402  5401 

 

统计f[i]出现次数即可

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int P[MAXN],siz=0,b[MAXN]={0};  
int gcd(int a,int b){if (b) return gcd(b,a%b);return a;}
void make_prime(int n)  
{  
    Fork(i,2,n)  
    {  
        if (!b[i])  
        {  
            P[++siz]=i;  
        }  
        For(j,siz)  
        {  
            if (P[j]*i>n) break;  
            b[P[j]*i]=b[i];
			if (gcd(P[j],i)==1) b[P[j]*i]++; 
            if (i%P[j]==0) break;  
        }  
    }  
}   
int t[8],cnt[MAXN][8]={0};
const int n=1000000,mm=7;
int main()
{
//	freopen("B.in","r",stdin);
	
	make_prime(n);
	For(i,n) 
	{
		For(j,7)
			cnt[i][j]=cnt[i-1][j]+(bool)(b[i]+1==j);
	}
	
	
//	For(i,30) cout<<i<<' '<<b[i]+1<<endl;	

	int T;cin>>T;
	while(T--)
	{
		int l,r;
		scanf("%d%d",&l,&r);
	
		For(i,7) t[i]=cnt[r][i]-cnt[l-1][i];	
		
		int ans=0;
		For(i,7)
			For(j,7)
			{
				if (t[i]&&t[j]&&((i!=j)||(t[i]>1))) ans=max(ans,gcd(i,j));
			}	
		printf("%d\n",ans);
		
	}
	return 0;
}