2015多校联合第三场 hdu5317 RGCDQ

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
   
   
   
   
   
   
   
   


   
   
   
   

这个题一看 ，问区间内最大的素因子个数，素数筛预处理+素因子分解果断超时 优化到不能再优化 依旧超时  how?

又是思维固化了orz  问素数个数不一定要求出素数啊 ~ init()函数用以求出每个数的素数种类 num[ ][ ]数组干嘛的 困扰了我一天 一重维度是每个数 另一重是每个数的质因子种类数 

什么跟什么啊这是 ￣□￣｜｜ 然后还把前一行的复制到下一行 下一行还有++ ……直到我枚举了前几个数

i\f[i]	1	2	3	4	5	6	7
2	1	0	0	0	0	0	0
3	2	0	0	0	0	0	0
4	3	0	0	0	0	0	0
5	4	0	0	0	0	0	0
6	4	1	0	0	0	0	0
7	5	1	0	0	0	0	0
8	6	1	0	0	0	0	0
9	7	1	0	0	0	0	0
10	7	2	0	0	0	0	0
11	8	2	0	0	0	0	0

其中有一个结论是10^6的范围内 某数的质因子种类不会超过7 （把前7个素数2、3、5、7、11、13……乘一起也超过10^6了）

num[ ][ ]表示截止到某数 有1、2、3、……个素因子种类的数有多少个 嗯嗯 求区间减一下就好了 

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long

using namespace std;

int f[1000005];
int num[1000005][8];
void init()
{
    for(int i = 2; i < 1000005; i++)
    {
        if(f[i] == 0)
        {
            f[i] = 1;
            for(int j = i + i; j < 1000005; j += i)
                f[j]++;
        }
    }
    for(int i = 2; i < 1000005; i++)
    {
        memcpy(num[i], num[i - 1], sizeof num[i - 1]);
        num[i][f[i]]++;
    }
}
int main()
{
    init();
    int T;
    while(~scanf("%d", &T))
    {
        while(T--)
        {
            int l, r;
            scanf("%d%d", &l, &r);
            int a[8] = {0};
            for(int i = 1; i < 8; i++)
                a[i] = num[r][i] - num[l - 1][i];
            int ans = 1;
            for(int i = 7; i > 0; i--)
                if(a[i] > 1)
            {
                ans = i;
                break;
            }
            if(a[4] && a[2])
                ans = max(ans, 2);
            if(a[6] && a[3])
                ans = max(ans, 3);
            printf("%d\n", ans);
        }
    }
    return 0;
}