toj_2861.cpp

Tianjin University Online Judge 的80多道题目 .

/*2861.   Divisors Time Limit: 1.0 Seconds   Memory Limit: 65536K
Total Runs: 374   Accepted Runs: 153


We define the function f(x) = the number of divisors of x. Given two integers a and b (a ≤ b), please calculate f(a) + f(a+1) + ... + f(b).
Input
Two integers a and b for each test case, 1 ≤ a ≤ b ≤ 231 - 1. The input is terminated by a line with a = b = 0.
Output
The value of f(a) + f(a+1) + ... + f(b).
Sample Input

9 12
1 2147483647
0 0

Sample Output

15
46475828386

Hint
For the first test case:
9 has 3 divisors: 1, 3, 9.
10 has 4 divisors: 1, 2, 5, 10.
11 has 2 divisors: 1, 11.
12 has 6 divisors: 1, 2, 3, 4, 6, 12.
So the answer is 3 + 4 + 2 + 6 = 15.


Source: TJU Team Selection Contest 1*/
#include<cstdio>
#include<cmath>

int main()
{
  int a , b , i , j , k , sqrtI;
  long long result;
  while ( scanf( "%d%d" , &a , &b ) != EOF && a != 0 && b != 0 )
    {
      result = 0;
      for ( i = a; i <= b; i++ )
	{
	  result += 2;
	  sqrtI = static_cast< int >( sqrt( i ) );
	  for ( j = 2; j <= sqrtI; j++ )
	    {
	      if ( i % j == 0 )
		result += 2;
	    }
	  if ( sqrtI * sqrtI == i )
	    result --;
	}
      printf( "%ld\n" , result );
    }
  return 0;
}