NUC1077 Humble Numbers【数学计算+打表】

Humble Numbers

时间限制: 1000ms 内存限制: 65536KB

通过次数: 1总提交次数: 1

问题描述

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.

输入描述

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.

输出描述

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.

样例输入

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

样例输出

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.

来源

{Ulm Local 1996}

问题分析：（略）

这个问题和《POJ2247 HDU1058 UVA443 ZOJ1095 Humble Numbers【数学计算+打表】》是同一个问题，代码拿过来用就AC了。

程序说明：参见参考链接。

参考链接：POJ2247 HDU1058 UVA443 ZOJ1095 Humble Numbers【数学计算+打表】

题记：程序做多了，不定哪天遇见似曾相识的。

AC的C++程序如下：

/* POJ2247 HDU1058 UVA443 ZOJ1095 Humble Numbers */

#include <iostream>
#include <stdio.h>

using namespace std;

const int N = 5842;
int humble[N];

void maketable(int n)
{
    int i2, i3, i5, i7;
    int h2, h3, h5, h7;

    humble[0] = 1;
    i2 = i3 = i5 = i7 = 0;
    h2 = humble[i2] * 2;
    h3 = humble[i3] * 3;
    h5 = humble[i5] * 5;
    h7 = humble[i7] * 7;
    for(int i=1; i<n; i++) {
        humble[i] = min(min(h2, h3), min(h5, h7));

        if(h2 == humble[i]) {
            h2 = humble[++i2] * 2;
        }
        if(h3 == humble[i]) {
            h3 = humble[++i3] * 3;
        }
        if(h5 == humble[i]) {
            h5 = humble[++i5] * 5;
        }
        if(h7 == humble[i]) {
            h7 = humble[++i7] * 7;
        }
    }
}

int main()
{
    maketable(N);

    int n;
    while(cin >> 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.
", humble[n - 1]);
    }

    return 0;
}