zoukankan      html  css  js  c++  java
  • HDU1058 Humble Numbers 【数论】

    Humble Numbers

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 17407    Accepted Submission(s): 7565


    Problem 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.

    #include <stdio.h>
    #define maxn 5845
    
    int dp[maxn] = {1, 1};
    int a, b, c, d;
    const char *sam[] = { "st", "nd", "rd", "th" };
    
    int min(int u, int v, int x, int y)
    {
        int tmp = u;
        if(tmp > v) tmp = v;
        if(tmp > x) tmp = x;
        if(tmp > y) tmp = y;
    
        if(tmp == u) ++a;
        if(tmp == v) ++b;
        if(tmp == x) ++c;
        if(tmp == y) ++d;
        
        return tmp;
    }
    
    const char *f(int n)
    {
        if(n % 10 == 1 && n % 100 != 11) return sam[0];
        if(n % 10 == 2 && n % 100 != 12) return sam[1];
        if(n % 10 == 3 && n % 100 != 13) return sam[2];
        return sam[3];
    }
    
    int main()
    {
        int i, n;
        a = b = c = d = 1;
        for(i = 2; i < maxn; ++i)
            dp[i] = min(dp[a] * 2, dp[b] * 3, dp[c] * 5, dp[d] * 7);
        while(scanf("%d", &n), n)
            printf("The %d%s humble number is %d.
    ", n, f(n), dp[n]);
        return 0;
    }


  • 相关阅读:
    设计模式~门面模式
    设计模式~享元模式
    设计模式~代理模式
    Java IO流:(六)节点流(文件流)之 FileWriter
    Java IO流:(五)节点流(文件流)之 FileReader
    Java IO流:(四)节点流(文件流)
    Java IO流:(三)IO流四大基类
    Java IO流:(二)IO流原理及流的分类
    Java IO流:(一)File 文件类
    每周总结
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/4510338.html
Copyright © 2011-2022 走看看