练习三1004humble number

最新推荐文章于 2019-06-01 16:57:39 发布

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最新推荐文章于 2019-06-01 16:57:39 发布

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本文介绍了一种特殊的数——谦逊数，这些数仅由2、3、5或7作为质因数组成。通过编程实现了一个算法，用于找出给定位置的谦逊数，并正确地输出其序数形式。

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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

题意：

st代表余数为1，nd代表余数为3，rd代表余数为5，特殊情况为11，12，13，。。th代表其余情况。

思路：

就是先求出humble number来，前者可表示为pow(2,a)*pow（3，b）*pow（5，c）*pow（7，d）的形式，可以先求出（0—5284）的所有humble number（他们都叫做打表

），根据从小到大的顺序排好序，其中t1,t2,t3,t4是用来排序的，记录了2，3，5，7，乘到几了。

代码:

#include<algorithm>
#include<stdlib.h>
#include<cstdio>
#include<string.h>
using namespace std;
int ary[6000];
int n;
int cmp(int a,int b,int c,int d)
{
    int temp;
    temp=min(a,b);
    temp=min(temp,c);
    temp=min(temp,d);
    return temp;
}
int main()
{
    int i;
    memset(ary,0,sizeof(ary));
    ary[1]=1;
    int t1=1,t2=1,t3=1,t4=1;
    for(i=2;i<=5842;i++)
    {
        ary[i]=cmp(2*ary[t1],3*ary[t2],5*ary[t3],7*ary[t4]);
        if(ary[i]==2*ary[t1]) t1++;
        if(ary[i]==3*ary[t2]) t2++;
        if(ary[i]==5*ary[t3]) t3++;
        if(ary[i]==7*ary[t4]) t4++;

    }
    while(cin>>n)
    {
        if(n<0||n>=5843)
        {
            cout<<"越界"<<endl;
            abort();
        }
        printf("The %d",n);
        if(n%10==1&&n%100!=11)  printf("st ");
        if(n%10==2&&n%100!=12)  cout<<"nd ";
        if(n%10==3&&n%100!=13)  printf("rd ");
        else printf("th ");
        printf("humble number is %d.\n",ary[n]);
    }
    return 0;
}

感想：

以前看着难，现在就是水题。。