CodeForces - 546D Soldier and Number Game【素数筛+前缀和】

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本文介绍了一种通过筛选素数并计算因子数量的方法来解决一个关于游戏策略的问题。该问题涉及两个士兵轮流选择正整数因子，直到无法再选择为止。文章详细说明了如何使用素数筛法找到特定范围内的所有素数，并计算每个数可以分解的因子数量。通过这种方法，可以确定初始值为a!/b!时，游戏能够进行的最大轮数。

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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
inputCopy
2
3 1
6 3
outputCopy
2
5

题意：

两个人每轮取n的一个因子k，将n/k交给另一个人，n初始值为a!/b!，n为1时结束。求可以进行多少轮。

思路：

可以利用合适的素数筛筛选出MAXN范围内的素数，同时可以计算每一个数可以分解的因子数。然后利用前缀和计算出来前a个数的因子数和，计算时用dp[a]-dp[b]即可。

代码：

#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define inf 0x3f3f3f3f
const int MAXN = 5e6;
int n,m,a[MAXN+5];
int num[MAXN+5],cnt[MAXN+5],dp[MAXN+5];
bool is_prime[MAXN+5];
void shai()
{
    memset(is_prime,1,sizeof is_prime);
    memset(cnt,0,sizeof cnt);
    for(int i=1;i<=MAXN;i++)
        num[i]=i;
    for(int i=2;i<=MAXN;i++)
    {
        if(is_prime[i])
        {
            cnt[i]++;
            for(int j=i+i;j<=MAXN;j+=i)
            {
                while(num[j]%i==0)
                {
                    num[j]/=i;
                    cnt[j]++;
                }
                is_prime[j]=0;
            }
        }
    }
}
int main()
{
    memset(dp,0,sizeof dp);
    shai();
    for(int i=1;i<=MAXN;i++)
        dp[i]=dp[i-1]+cnt[i];
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",dp[a]-dp[b]);
    }
    return 0;
}