CodeForces 546D Soldier and Number Game(素数筛选)

http://codeforces.com/problemset/problem/546/D

Soldier and Number Game

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

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.

Sample test(s)

input

23 16 3

output

25

题意：

给出一个n，n开始是a!/b!，每次用一个x去整除n得到新的n，最后当n变成1的时候经过了几轮得分就是这个轮数，要求最大的分数是多少

思路：

很明显，就是一个求整数质因子个数的题目，阶乘我们不需要算，我们知道在a>b的时候，b!都约掉了，那么我们只需压迫计算出每个数的质因数有几个，然后计算出1~n的质因子之和，那么就可以迅速得到答案了

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <cstdlib>#include <limits>#include <queue>#include <stack>#include <vector>#include <map>using namespace std;typedef long long LL;#define N 5100000#define INF 0x3f3f3f3f#define PI acos (-1.0)#define EPS 1e-8#define met(a, b) memset (a, b, sizeof (a))int sum[N], isprim[N]={1, 1, 0};void Init (){    for (int i=2; i<=N; i++)    {        if (!isprim[i])        {            for (int j=i; j<=N; j+=i)            {                int a = j;                while (a % i == 0)                {                    sum[j]++;                    a /= i;                }                isprim[j] = 1;            }        }    }    for (int i=1; i<=N; i++)        sum[i] += sum[i-1];}int main (){    int t;    scanf ("%d", &t);    Init ();    while (t--)    {        int a, b;        scanf ("%d %d", &a, &b);        printf ("%d\n", sum[a]-sum[b]);    }    return 0;}