hdu 5317 合数分解+预处理

RGCDQ

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2818    Accepted Submission(s): 1108

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

/*
hdu 5317 合数分解+预处理

problem:
查找区间[l,r]中 gcd(F[a[i]],F[a[j]])的最大值. F[x]为x的分解出的质因子种类数

solve:
可以先计算一下,1e6时质因子最多有7个. 所以可以dp[maxn][7]先预处理出质因子个数的前缀和.
然后查找 1~7谁出现了2次及以上

hhh-2016-08-21 10:38:45
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson  ch[0]
#define rson  ch[1]
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 1000000;
const int INF = 1e9+10;

int prime[maxn+1];

void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2;i <= maxn;i++)
    {
        if(!prime[i]) prime[++prime[0]] = i;
        for(int j = 1;j <= prime[0] && prime[j] <= maxn/i;j++)
        {
            prime[prime[j]*i] = 1;
            if(i % prime[j] == 0) break;
        }
    }
}

int getFactor(int x)
{
    int t = x;
    int fant = 0;
    for(int i = 1;prime[i] <= t/prime[i];i++)
    {
        if(t % prime[i] == 0)
        {
            fant ++;
            while(t % prime[i] == 0)
                t /= prime[i];
        }
    }
    if(t != 1)
        fant ++;
    return fant;
}

int dp[maxn+1][7];

int main()
{
    getPrime();
    for(int i = 0;i <= 7;i++)
        dp[0][i] = 0;
    for(int i = 1;i <= maxn;i++)
    {
        int t = getFactor(i);
        for(int j = 0;j < 7;j++)
        {
            if(t == j+1)
                dp[i][j] = dp[i-1][j] + 1;
            else
                dp[i][j] = dp[i-1][j];
        }
    }
    int T;
    int a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&a,&b);
        int tMax = 0;
        for(int i = 6;i >= 0;i--)
        {
            if(dp[b][i] - dp[a-1][i] > 1)
            {
                tMax =i;
                break;
            }
        }
        printf("%d\n",tMax+1);
    }
}