HDU 1406 Completion (number theory)

After the number

http://acm.hdu.edu.cn/showproblem.php?pid=1406

Time limit:2000/1000 MS (java/others)

Memory limit:65536/32768 K (java/others)

Problem Description

The definition of a number: if the sum of all the factors of a positive integer greater than 1 equals its own, it is said that the number is the end of the number, such as 6, 28 is the end: 6=1+2+3;28=1+2+4+7+14. The task of the subject is to judge the number of the number of two positive integers.

Input

The input data contains multiple rows, the first row is a positive integer n, which indicates the number of test instances, then n test instances, one row for each instance, consisting of two positive integers num1 and num2 (1<num1,num2<10000).

This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45516.htm

Output

For each set of test data, output the number of NUM1 and num2 (including NUM1 and num2).

Sample Input

22 55 7

Sample Output

01

Note the interval endpoint problem.

Complete code:

/*0ms,208kb*/
    
#include <cstdio>  
#include <algorithm>  
using namespace std;  
const int perfect_number[] = {6, 496, 8128};  
    
int main ()  
{  
    int T, a, B, I, J;  
    scanf ("%d", &t);  
    while (t--)  
    {  
        scanf ("%d%d", &a, &b);  
        if (a > B) Swap (A, b);  
        for (i = 0; i < 4; ++i)  
            if (a <= perfect_number[i]) break;  
        for (j = 0; j < 4; ++j)  
            if (b < perfect_number[j]) break;///Be careful!  
        printf ("%d\n", j-i);  
    }  
    return 0;  
}