CodeForces - 546D Soldier and Number Game（素数筛法打表前缀和）

output

standard output

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

Copy

2
3 1
6 3

output

Copy

2
5

这是一道数论题，以前做过一个和这个差不多的，不过没写过博客，记录一发。

题意：给你啊a,b。求出从a，a-1,a-2，...，b+1范围内内个数的素数个数和。例如a=6,b=3;那么6就是2*3,5就是5,4是2*2

那么个数就是2+1+2=5；

那么我们可以用素数筛法先求出每个数有多少个素数，然后再求一下前缀和就能得到结果。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#define inf 0x3f3f3f3f
#define ll long long
#define maxx 5000000
using namespace std;
int ans[5000005];
int prime[5000005];
void pp()
{
    for(int i=2; i<=5000005; i++)
    {
        if(prime[i]==0)
        {
            for(int j=i; j<=5000005; j+=i)
            {
                int res=j;
                while(res%i==0)
                {
                    ans[j]++;
                    res/=i;
                }
                prime[j]=1;
            }
        }
    }
}
int main()
{
    memset(prime,0,sizeof(prime));
    memset(ans,0,sizeof(ans));
    pp();
    for(int i=2;i<=5000005;i++)
        ans[i]=ans[i-1]+ans[i];
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int a, b;
        scanf("%d%d",&a,&b);
        printf("%d\n",ans[a]-ans[b]);
    }
    return 0;
}