• Humble Numbers


    Humble Numbers
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 9988 Accepted: 4665

    Description
    A number whose only prime factors are 2,3,5 or 7 is called a humble number. The sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, … shows the first 20 humble numbers.

    Write a program to find and print the nth element in this sequence.

    Input
    The input consists of one or more test cases. Each test case consists of one integer n with 1 <= n <= 5842. Input is terminated by a value of zero (0) for n.

    Output
    For each test case, print one line saying “The nth humble number is number.”. Depending on the value of n, the correct suffix “st”, “nd”, “rd”, or “th” for the ordinal number nth has to be used like it is shown in the sample output.

    Sample Input

    1
    2
    3
    4
    11
    12
    13
    21
    22
    23
    100
    1000
    5842
    0

    Sample Output

    The 1st humble number is 1.
    The 2nd humble number is 2.
    The 3rd humble number is 3.
    The 4th humble number is 4.
    The 11th humble number is 12.
    The 12th humble number is 14.
    The 13th humble number is 15.
    The 21st humble number is 28.
    The 22nd humble number is 30.
    The 23rd humble number is 32.
    The 100th humble number is 450.
    The 1000th humble number is 385875.
    The 5842nd humble number is 2000000000.

    Source
    Ulm Local 1996
    好恶心的输出格式

    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <string>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #define LL long long
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    
    const LL MAX = 2000000000;
    
    LL A[10000];
    int top;
    int main()
    {
        //freopen("output.txt","w",stdout);
        LL a,b,c,d;
        top=0;
        for(int i=0; i<=33; i++)
        {
            if(i==0)
            {
                a=1;
            }
            else
            {
                a*=2;
            }
            if(a>MAX)
            {
                break;
            }
            for(int j=0; j<=33; j++)
            {
                if(j==0)
                {
                    b=1;
                }
                else
                {
                    b*=3;
                }
                if(a*b>MAX)
                {
                    break;
                }
                for(int k=0; k<=33; k++)
                {
                    if(k==0)
                    {
                        c=1;
                    }
                    else
                    {
                        c*=5;
                    }
                    if(a*b*c>MAX)
                    {
                        break;
                    }
                    for(int s=0; s<=33; s++)
                    {
                        if(s==0)
                        {
                            d=1;
                        }
                        else
                        {
                            d*=7;
                        }
                        if(a*b*c*d>MAX)
                        {
                            break;
                        }
                        else
                        {
                            A[top++]=a*b*c*d;
                        }
                    }
                }
            }
        }
        sort(A,A+top);
        int n;
        while(scanf("%d",&n)&&n)
        {
            if(n%10==1&&(n%100>=20||n%100<10))
            {
                printf("The %dst humble number is %I64d.
    ",n,A[n-1]);
            }
            else if(n%10==2&&(n%100>=20||n%100<10))
            {
                printf("The %dnd humble number is %I64d.
    ",n,A[n-1]);
            }
            else if(n%10==3&&(n%100>=20||n%100<10))
            {
                printf("The %drd humble number is %I64d.
    ",n,A[n-1]);
            }
            else
            {
                printf("The %dth humble number is %I64d.
    ",n,A[n-1]);
            }
        }
        return 0;
    }
    
  • 相关阅读:
    [SDOI2013]直径(树的直径)
    [ZJOI2012]旅游(树的直径)
    [SDOI2011]消防(树的直径)
    【模板】2-SAT 问题(2-SAT)
    [HNOI2006]公路修建问题
    速度限制(分层图)
    [JLOI2011]飞行路线(分层图)
    【洛谷 P3194】 [HNOI2008]水平可见直线 (单调栈)
    【洛谷 P3187】 [HNOI2007]最小矩形覆盖 (二维凸包,旋转卡壳)
    【洛谷 P1452】 Beauty Contest (二维凸包,旋转卡壳)
  • 原文地址:https://www.cnblogs.com/juechen/p/5256002.html
Copyright © 2020-2023  润新知