HDU 2521 Inverse prime

Inverse Prime Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 4142 Accepted Submission (s): 2396


Problem Description The inverse prime is satisfied for any I (0<i<x), there is g (i) <g (x), (g (x) is the number of factors of x), then x is an inverse prime. Now give you an integer interval [a, b], please find out the x of the interval to make g (x) the largest.

Input first line n, next n line test data
The input includes a, B, 1<=a<=b<=5000, which indicates the closed interval [a, b].

The output is an integer that is the maximum number of the interval factor. If there are more than one, the minimum number is output.

Sample Input

3 2 3 1 10 47 359
Sample Output

2 6 240

#include <stdio.h> 
int s[5010]={0,1,2,2};
void f ()
{
    int i,j;
    for (i=4;i<5010;i++)
    {for
        (j=2;j<=i/2;j++)//I/2 represents greatest common divisor 
        //for (j=2;j*j<=i;j++)
        if (i%j==0 )  s[i]++;
        s[i]+=2;
    }
}
int main ()
{
	f ();
    int n,a,b,i,max,t;
    scanf ("%d", &n);
    while (n--)
    {
        scanf ("%d%d", &a,&b);
        T=a;
        max=0;
        for (i=a;i<=b;i++)
        {
            if (S[i]>max) 
            {  
                max=s[i];
                t=i;
            }
        }
        printf ("%d\n", t);
    }
    return 0;
}