D. Soldier and Number Game

D. Soldier and Number Gametime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputTwo soldiers are playing a game. At the beginning first of them chooses a positive integer n and gives it to the second soldier. Thenthe second one tries to make maximum possible number of rounds. Each round consists of choosing a positive integer x > 1, such that n isdivisible by x and replacing n with n / x.When n becomes equal to 1 andthere 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! forsome positive integer a and b (a ≥ b).Here by k! we denote the factorial of k thatis defined as a product of all positive integers not large than k.What is the maximum possible score of the second soldier?InputFirst 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.OutputFor each game output a maximum score that the second soldier can get.Examplesinput

2
3 1
6 3

output

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;}