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最新推荐文章于 2020-03-16 11:43:13 发布

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最新推荐文章于 2020-03-16 11:43:13 发布

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本文探讨了一种特殊的游戏策略，其中第一玩家选择形式为 a! / b! 的整数 n，第二玩家通过最大化的回合数来获取分数。文章提供了一个算法实现方案，包括预处理阶段和输入处理阶段，以快速计算出每组 (a, b) 对应的最大可能得分。

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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.

Sample test(s)

input
2
3 1
6 3

output
2
5

从今往后。。。我再也不信任CF的机器了。。竟然卡iostream。。。我的天

#include <bits/stdc++.h>
using namespace std;
int ps[5000010];
int main()
{
	for(int i=2;i<5000000;i++)
		if(ps[i]==0)
			for(int j=i;j<=5000000;j+=i)
				ps[j]=ps[j/i]+1;
	for(int i=1;i<=5000000;i++)
		ps[i]+=ps[i-1];
	int n;
	cin>>n;
	while(n--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		printf("%d\n",ps[a]-ps[b]);
	}
}