Description 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≤1000000) denoting number of games soldiers play. Then follow t lines, each contains pair of integers a and b (1≤b≤a≤5000000) defining the value of n for a game. Output
For each game output a maximum score that the second soldier can get.
Sample Input
Input 2 3 1 6 3 Output 2 5
题解：就是让求a!/b!的质因子个数。用到了素数筛。
<span style="font-family:SimSun;">#include<cstdio>
#include<iostream>
using namespace std;
#define N 5000100
int vis[N],s[N];
int main()
{
fill(s,s+N,0);
fill(vis,vis+N,0);
for(int i=2;i<=N;i++)
{
if(!vis[i])
{
for(int j=i;j<=N;j+=i)
{
int k=j;
while(k%i==0)
{
s[j]++;
k/=i;
}
vis[j]=1;
}
}
}
for(int i=2;i<=N;i++)
s[i]+=s[i-1]; //表示前 i个数因子个数之和
int t,a,b;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&a,&b);
printf("%d\n",s[a]-s[b]);
}
return 0;
} </span>
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