CodeForces - 546D Soldier and Number Game（素數篩法打表前綴和）

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

這是一道數論題，以前做過一個和這個差不多的，不過沒寫過博客，記錄一發。

題意：給你啊a,b。求出從a，a-1,a-2，...，b+1範圍內內個數的素數個數和。例如a=6,b=3;那麼6就是2*3,5就是5,4是2*2

那麼個數就是2+1+2=5；

那麼我們可以用素數篩法先求出每個數有多少個素數，然後再求一下前綴和就能得到結果。

#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;
}