codeforce Round304 div.2 D

题意：

Two soldiers are playing a game. At the beginning first of them chooses a positive integer
n and gives it to the second soldier. Then the second one tries to make maximum possible number of rounds. Each round consists of choosing a positive integer
x > 1, such that n is divisible by
x and replacing n with
n / x. When
n becomes equal to 1 and there is no more possible valid moves the game is over and the score of the second soldier is equal to the number of rounds he performed.

To make the game more interesting, first soldier chooses
n of form a! / b! for some positive integer
a and b (a ≥ b). Here by
k! we denote the
factorial of k that is defined as a product of all positive integers not large than
k.

What is the maximum possible score of the second soldier?

Input
First line of input consists of single integer t (1 ≤ t ≤ 1 000 000) denoting number of games soldiers play.

Then follow t lines, each contains pair of integers
a and b (1 ≤ b ≤ a ≤ 5 000 000) defining the value of
n for a game.

Output
For each game output a maximum score that the second soldier can get.

找出一个大范围内的每个数的因子的最大数目，并把他们加起来。很明显每个合数都可以分解为质数的乘积，所以找出质因子（可重复）的个数就行了。枚举从小到大的每个质数，把能乘除它的合数的质因子数都加一，最后就能得到每个数的质因子（可重复）个数。

两个注意点：

1.对于一个质数而言，一个能整除它的合数可能不止能整除它一次，所以要把它除尽，每次除法都把质因子个数加一

2.比赛的时候没有用枚举质数去求能整除它的合数，而是枚举合数去求因子个数，结果超时了。

代码：