Codeforces Round #304 (Div. 2) D. Soldier and Number Game（筛选素因子，前缀和）

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本文探讨了通过素数分解解决特定数学问题的方法，并应用于一个两士兵的游戏策略中。介绍了如何利用前缀和来高效计算一定范围内整数的最大素数因子数目，并分析了基于阶乘形式的数值在游戏中所能达到的最大轮数。

题目链接：点击打开链接

题意：两个整数a, b。求出a, a - 1, a - 2........b +1这些整数最多能被拆分成多少个数相乘。

思路：数量最多，当然都要拆成素数。因为是要查1e6次区间，所以自然能想到前缀和。要计算每个数最多拆成几个素因子，可以筛出来，然后再计算前缀和。

记得以前写过筛素因子...结果一时没能想起来...

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e6+5;
int sum[maxn];

void init()
{
    for(int i = 2; i < maxn; i++)
    {
        if(!sum[i])
            for(int j = i; j < maxn; j += i)
                sum[j] = sum[j/i]+1;
    }
    for(int i = 1; i < maxn; i++)
        sum[i] += sum[i-1];
}

int main(void)
{
    init();
    int t, a, b;
    cin >> t;
    while(t--)
    {
        scanf("%d%d", &a, &b);
        printf("%d\n", sum[a]-sum[b]);
    }
    return 0;
}

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.

        Examples
      
          input
        
2
3 1
6 3

          output
        
2
5