2015 Multi-University Training Contest 3 hdu 5317 RGCDQ

RGCDQ

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1652    Accepted Submission(s): 696

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

 

解题：打表。。。学习了Crazyacking的写法

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;

int sum[maxn][8],kinds[maxn];
bool np[maxn];
void init() {
    for(int i = 2; i < maxn; ++i)
        if(!np[i]) {
            for(int j = i; j < maxn; j += i) {
                np[j] = true;
                ++kinds[j];
            }
        }
    for(int i = 1; i < maxn; ++i)
        for(int j = 7; j > 0; --j)
            sum[i][j] = kinds[i] == j;
    for(int i = 1; i < maxn; ++i)
        for(int j = 7; j > 0; --j)
            sum[i][j] += sum[i-1][j];
    //记录1到上界含有j个不同素因子的数的个数
}
int main() {
    int kase,a,b;
    init();
    scanf("%d",&kase);
    while(kase--) {
        scanf("%d%d",&a,&b);
        for(int i = 7; i > 0; --i) {
            if(sum[b][i] - sum[a-1][i] > 1) {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}