zoukankan      html  css  js  c++  java
  • codeforces Soldier and Number Game(dp+素数筛选)

                                  D. Soldier and Number Game
                              time limit per test3 seconds
                              memory limit per test256 megabytes
                              inputstandard input
                              outputstandard 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
    2
    3 1
    6 3
    output
    2
    5
    

      

    /*
        dp求解 n!的质因子个数。 
        1.首先利用素数筛选法求出素数的集合
        2.dp[i] = dp[i/prime[j]]+1, i%prime[j]==0; prime[j]是第一个能整除i的数 
    */
    #include<iostream> 
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    #define N 5000005
    using namespace std;
    
    int prime[N];
    bool isprime[N];
    int dp[N];
    void init(){
        int top = 0;
        for(int i=2; i<N; ++i){
            if(!isprime[i])
                prime[top++] = i;
            for(int j=0; j<top && i*prime[j]<N; ++j){//素数筛选 
                isprime[i*prime[j]] = true;
                if(i%prime[j] == 0)
                    break;
            }
        }
        for(int i=2; i<N; ++i){
            if(!isprime[i])
                dp[i] = 1;
            else{
                for(int j=0; j<top; ++j)
                    if(i%prime[j]==0){
                        dp[i] = dp[i/prime[j]] + 1;
                        break;
                    }
            }
        }
        for(int i=2; i<N; ++i)//前n个数的质因数个数的和 
            dp[i]+=dp[i-1];
    }
     
    int main(){
        int t;
        scanf("%d", &t);
        init();
        while(t--){
            int a, b;
            scanf("%d%d", &a, &b);
            printf("%d
    ", dp[a] - dp[b]);
        }
        return 0; 
    } 
  • 相关阅读:
    elementUI .native修饰符
    JS密码强度检测
    HighChart中的tooltip的第一行数字明显比其他的字要小
    HighChart 不同颜色(柱状图)
    .NET(C#、VB)APP开发——Smobiler平台控件介绍:SliderView控件
    .NET(C#、VB)APP开发——Smobiler平台控件介绍:SignatureButton控件
    .NET(WinCE、WM)转Android开发——Xamarin和Smobiler对比
    MobileForm控件的使用方式-用.NET(C#)开发APP的学习日志
    你用.NET开发APP时,在云平台打包APP要填个“包名”的含义
    基于.NET的APP开发和Windows开发,异步回调差别
  • 原文地址:https://www.cnblogs.com/hujunzheng/p/4529072.html
Copyright © 2011-2022 走看看