zoukankan      html  css  js  c++  java
  • Diophantus of Alexandria

    Diophantus of Alexandria was an egypt mathematician living in Alexandria. He was one of the first mathematicians to study equations where variables were restricted to integral values. In honor of him, these equations are commonly called diophantine equations. One of the most famous diophantine equation is x^n + y^n = z^n. Fermat suggested that for n > 2, there are no solutions with positive integral values for x, y and z. A proof of this theorem (called Fermat's last theorem) was found only recently by Andrew Wiles. 

    Consider the following diophantine equation: 

    1 / x + 1 / y = 1 / n where x, y, n ∈ N+ (1)



    Diophantus is interested in the following question: for a given n, how many distinct solutions (i. e., solutions satisfying x ≤ y) does equation (1) have? For example, for n = 4, there are exactly three distinct solutions: 

    1 / 5 + 1 / 20 = 1 / 4 
    1 / 6 + 1 / 12 = 1 / 4 
    1 / 8 + 1 / 8 = 1 / 4




    Clearly, enumerating these solutions can become tedious for bigger values of n. Can you help Diophantus compute the number of distinct solutions for big values of n quickly? 

    Input

    The first line contains the number of scenarios. Each scenario consists of one line containing a single number n (1 ≤ n ≤ 10^9). 

    Output

    The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Next, print a single line with the number of distinct solutions of equation (1) for the given value of n. Terminate each scenario with a blank line. 

    Sample Input

    2
    4
    1260

    Sample Output

    Scenario #1:
    3
    
    Scenario #2:
    113

    代码:

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    #define M 50005
    int  prime[50005];
    void db()
    {
        int i,j;
        memset(prime,0,sizeof(prime));
        for(i=2; i<=M; i++)
        {   
            if(prime[i]==0)
            {
                for(j=i+i; j<=M; j+=i)
                {
                     prime[j]=1;
                }
            }
        }
    }
    int main()
    {
    
        db();
        int n,i,j,k,t;
        scanf("%d",&t);
        int sum;
        int cnt=1;
        while(t--)
        {
            sum=1;
           
            scanf("%d",&n);
            for(i=2; i<=M; i++)
            {
                if(n==1)
                break;
                if(prime[i]==0)
                {
                        k=0;
                        while(n%i==0)
                        {
                            k++;
                            n=n/i;
                        }
                        sum=sum*(2*k+1);
                }
            }
    
            if(n>1)
                sum=sum*3;
            printf("Scenario #%d:
    ",cnt);
            printf("%d
    
    ",(sum+1)/2);
            cnt++;
    
        }
        return 0;
    
    }
  • 相关阅读:
    less的一些用法整理
    flex 布局下关于容器内成员 flex属性的理解
    Web开发者需具备的8个好习惯
    GridView分页功能的实现
    程序员长期保持身心健康的几点建议
    程序员必知的10大基础实用算法
    结对编程
    Python_Day_02 str内部方法总结
    Python_Day_01(使用环境为Python3.0+)
    圆形头像制作
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10782083.html
Copyright © 2011-2022 走看看