HDU 5317（2015多校第三场1002）

RGCDQ

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1320    Accepted Submission(s): 573

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

 

题目大意：

给定L,R，求MAX( gcd ( f[i] , f[j] ) ) L<=i,j<=R

而 f（x）的值为 x的不同素数因子的个数 其中x<=1000000.

因此我们可以知道，f（x） 最大为 7 。 （2*3*5*7*11*13*17即等于510510）

因此我们可知答案只可能为 1-7中的数字

我们只要知道了在L,R中1-7中的数字分别出现了几次，我们便能求出最大的gcd值。

这个则可以用类似求前缀和的方法进行一次预处理

f（x）则可以用素数的筛选法来求解。（比赛一开始时错误的估计了该方法的时间复杂度，导致一开始的时候愣是不敢下手。。。。。。。。）

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <sstream>
#define PI acos(-1.0)
#define eps 1e-8
#define f 1000010
const int  inf =  (1<<30) - 10;
using namespace std;

int s[f], p;
int num[f];
int dp[f][8];

void get_prime()
{
	int i,j,n,m,s1,s2,a,b;
	s[1]=2;
	s[2]=3;
	m=2;
	for(i=6;i<f;i=i+6)
	{
		a=i-1;
		b=i+1;
		s1=1;
		s2=1;
		n=(int)sqrt(m);
		for(j=1;j<=n+1;j++)
			if(a%s[j]==0){s1=0;break;}
		for(j=1;j<=n+1;j++)
			if(b%s[j]==0){s2=0;break;}
		if(s1!=0){m++;s[m]=a;}
		if(s2!=0){m++;s[m]=b;}
	}
	p = m;
}

void init(){
    memset(num,0,sizeof(num));
    memset(dp,0,sizeof(dp));
    for(int i = 1;i <= p; ++i){
        for(int j = 1;j*s[i] <= 1000000; ++j){
            num[j*s[i]]++;
        }
    }
    for(int i = 2;i <= 1000000; ++i){
        for(int j = 1;j <= 7; ++j){
            dp[i][j] = dp[i-1][j];
        }
        dp[i][num[i]]++;
    }
}

int main(){
    //freopen("input.txt","r",stdin);
    get_prime();
    init();
    int T;
    cin>>T;
    int l, r;
    while(T--){
        scanf("%d %d",&l,&r);
        int knum[8];
        for(int i = 1;i <= 7; ++i){
            knum[i] = dp[r][i] - dp[l-1][i];
        }
        if(knum[7] >= 2){
            printf("7\n");
        }else if(knum[6]>=2){
            printf("6\n");
        }else if(knum[5]>=2){
            printf("5\n");
        }else if(knum[4]>=2){
            printf("4\n");
        }else if(knum[3]>=2){
            printf("3\n");
        }else if(knum[6] && knum[3]){
            printf("3\n");
        }else if(knum[2]>=2){
            printf("2\n");
        }else if(knum[6] && knum[2]){
            printf("2\n");
        }else if(knum[4] && knum[2]){
            printf("2\n");
        }else{
            printf("1\n");
        }
    }
    return 0;
}

如有BUG，请大家务必指出，不胜感激~

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