POJ2247 HDU1058 UVA443 ZOJ1095 Humble Numbers【数学计算+打表】

Humble Numbers

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10731 Accepted: 5015

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

1234111213212223100100058420

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

问题链接：POJ2247 HDU1058 UVA443 ZOJ1095 Humble Numbers。

问题简述：参见上文。

问题分析：

计算谦虚数问题，关键在于如何从小到大计算谦虚数。

根据题意，1是谦虚数；谦虚数是2,3,5和7的倍数；最小的谦虚数乘以2,3,5和7是谦虚数。

计算过程的关键是能够按照从小到大顺序算出谦虚数。

打表是必要的，可以避免重复计算。

计算过程中也需要尽量减少计算量。

程序说明：（略）

参考链接：（略）

题记：（略）

AC的C++语言程序如下：

/* 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.\n", humble[n - 1]);    }    return 0;}