hdu 1058 dp.Humble Numbers

最新推荐文章于 2018-07-19 23:44:21 发布

weixin_30834783

最新推荐文章于 2018-07-19 23:44:21 发布

阅读量74

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/nyist-xsk/p/7264904.html

本文介绍了一种特殊的数列——谦逊数（仅包含2、3、5、7为质因数的数），并提供了一个算法实现，用于找出该数列中的第n个元素。文章通过C++代码详细展示了如何高效生成这些数，并处理了输出格式中关于序数词的细节。

Humble Numbers

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1058

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

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 <iostream>
#include <stdio.h>
using namespace std;
int f[5843],n;
int i,j,k,l;

int min(int a,int b,int c,int d)
{
    int min=a;
    if(b<min) min=b;
    if(c<min) min=c;
    if(d<min) min=d;

    if(a==min) i++;
    if(b==min) j++;
    if(c==min) k++;
    if(d==min) l++;

    return min;//a或b或c或d
}

int main()
{
    i=j=k=l=1;
    f[1]=1;
    for(int t=2; t<=5842; t++)
    {
        f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);
    }
    while(scanf("%d",&n)&&n!=0)
    {
        if(n%10==1&&n%100!=11)
            printf("The %dst humble number is %d.\n",n,f[n]);
        else if(n%10==2&&n%100!=12)
            printf("The %dnd humble number is %d.\n",n,f[n]);
        else if(n%10==3&&n%100!=13)
            printf("The %drd humble number is %d.\n",n,f[n]);
        else
            printf("The %dth humble number is %d.\n",n,f[n]);
    }
    return 1;
}



//#include<stdio.h>
//#define min(a,b) (a<b?a:b)
//#define min4(a,b,c,d) min(min(a,b),min(c,d))
//int a[5850];//存放丑数
//
//int main()
//{
//    int n=1;
//    int p2,p3,p5,p7;
//    p2=p3=p5=p7=1;//2,3,5,7的计数器
//    a[1]=1;
//    while(a[n]<2000000000)//从2开始递推计算，一共5842个丑数
//    {
//        a[++n] = min4(2*a[p2],3*a[p3],5*a[p5],7*a[p7]);//取最小值，相应的计数器加1
//        if(a[n]==2*a[p2]) p2++;
//        if(a[n]==3*a[p3]) p3++;
//        if(a[n]==5*a[p5]) p5++;
//        if(a[n]==7*a[p7]) p7++;
//    }
//    while(scanf("%d",&n) && n)
//    {
//        if(n%10 == 1&&n%100!=11)
//            printf("The %dst humble number is ",n);
//        else if(n%10 == 2&&n%100!=12)
//            printf("The %dnd humble number is ",n);
//        else if(n%10 == 3&&n%100!=13)
//            printf("The %drd humble number is ",n);
//        else
//            printf("The %dth humble number is ",n);
//        printf("%d.\n",a[n]);
//    }
//    return 0;
//
//}

转载于:https://www.cnblogs.com/nyist-xsk/p/7264904.html