Because the exchange of several wrong .... And the pit point is num1 can be greater than num2

Finish number

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 22289 Accepted Submission (s): 8150

Problem description Definition: If a positive integer greater than 1 equals the sum of all the factors equal to its own, then the number is said to be the number, such as 6, 28 is the completion of the number: 6=1+2+3;28=1+2+4+7+14.

The task is to determine the number of two positive integers.

Input data contains multiple lines, the first line is a positive integer n, which indicates the number of test instances, then n test instances, each of which consists of two positive integers num1 and num2, (1<num1,num2<10000).

Output for each set of test data, print out the number of NUM1 and num2 (including NUM1 and num2) that exist.

Sample Input22 55 7

Sample Output01

1#include <iostream>2#include <cstdio>3 using namespacestd;4 int_is (intx)5 {6     inti,j,k,l,sum=1;7      for(i=2; i<=x/2; i++)8     {9         if(x%i==0)Ten         { Onesum+=i; A         } -     } -     if(sum==x) the         return 1; -     return 0; - } - intMain () + { -     intX,y,i,j,n; +Cin>>N; A      while(n--) at     { -         intCount=0; -scanf"%d%d",&x,&y); -         if(x>y) -         { -             intT; int=x; -x=y; toy=T; +         } -          for(i=x;i<=y;i++) the         { *             if(_is (i)) $             {Panax Notoginsengcount++; -             } the         } +printf"%d\n", count); A     } the      +}

Because the exchange of several wrong .... And the pit point is num1 can be greater than num2