RGCDQ(筛法）

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

思路：利用筛法的特性求出2到一百万的每个数本身有几类素因子组成。然后，因为最多种类不会超过7,（2*3*5*7*11*13*17*19超过了

1000000），所以进行判断。另外，这道题用c++输入输出会超时。

代码：

#include
#include
#include
#include
#include
using namespace std;

typedef long long LL;
const int N=1000000;
bool a[N+10];
int b[N+10];
int s[N+10][8];
int t[8];
int main()
{
    int T;
    cin>>T;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    for(int i=2; i<=N; i++)
    {
        if(!a[i])
            for(int j=i; j<=N; j+=i)
            {
                a[j]=1;
                b[j]++;
            }
        for(int j=1; j<=7; j++)
            s[i][j]=s[i-1][j];
        s[i][b[i]]++;
    }
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int p=1;
        for(int i=1; i<=7; i++)
        {
            t[i]=s[m][i]-s[n-1][i];
            if(t[i]>=2)
                p=max(i,p);
            if((t[2]&&t[4])||(t[4]&&t[6])||(t[2]&&t[6]))
                p=max(p,2);
        }
        printf("%d\n",p);
    }
}