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?
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.
For each game output a maximum score that the second soldier can get.
2 3 1 6 3
2 5
#include <cstdio> #include <cstdlib> #include <iostream> #include <climits> using namespace std; typedef long long int ll; const int maxn = 5e6+5; int prime[maxn]; ll num[maxn]; void Init(){ for(int i=2; i<=5000000; i++){ if(prime[i]==0){ for(int j=i; j<=5000000; j+=i){ int tmp = j; while(tmp%i==0){ num[j]++; tmp /= i; } prime[j] = 1; } } } for(int i=1; i<=5000000; i++) num[i] = num[i]+num[i-1]; } int main(){ Init(); int T; scanf("%d",&T); int a,b; while(T--){ scanf("%d %d",&a,&b); printf("%I64d\n",num[a]-num[b]); } return 0; }

本文介绍了一个两士兵间的计数游戏:一方选择形式为 a! / b! 的正整数 n 开始游戏;另一方通过不断除以大于1的因子来最大化游戏轮数,直至 n 降为1。文章提供了一种高效算法,用于计算不同 a 和 b 值下第二位士兵可能获得的最大分数。
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