HDU1058Humble Numbers

Humble Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25555    Accepted Submission(s): 11268

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

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.题目大意为存在这样的humble number他们的质因数仅有2，3，5，7.首先分析，如果一个数是humble number，那么他们的质因数必定可以分解为这样的形式：2^a * 3^b * 5^c * 7^d,那么第i个humble number怎样求呢；显然dp[i+1]要么等于之前的一个humble number dp[f1] * 2,要么等于另一个dp[f2] * 3,或dp[f3] * 5,或dp[f4] * 7;那么初始条件应为，dp[1]=1;f1=f2=f3=f4=1;那么代码中dp[i] = Min(dp[f1] * 2, dp[f2] * 3, dp[f3] * 5, dp[f4] * 7);为何要取最小的一个，因为这个递推需要从小到大递推，由底向上递推，保证其连续性，即求出来的i，是当前应当求出来的，最小的humble number；而其中为什么又会出现f1++,f2++,f3++,f4++呢，这是由于，当由任意一个ft,(t=1,2,3,4)求出当前dp[i]后，若要递推之后新的dp[i],显然应当更新这样的ft;给出DP求解代码：
#include <stdio.h>int min(int a, int b){    return a > b ? b : a;}int Min(int a, int b, int c, int d){    return min(a, min(b, min(c, d)));}int main(){    int dp[5843];    dp[1] = 1;    int f1, f2, f3, f4;    f1 = f2 = f3 = f4 = 1;    for (int i = 2; i <= 5842; i++)    {        dp[i] = Min(dp[f1] * 2, dp[f2] * 3, dp[f3] * 5, dp[f4] * 7);        if (dp[i] == dp[f1] * 2)            f1++;        if (dp[i] == dp[f2] * 3)            f2++;        if (dp[i] == dp[f3] * 5)            f3++;        if (dp[i] == dp[f4] * 7)            f4++;    }    int n;    while (scanf("%d", &n) != EOF)    {        if (n == 0)            break;        printf("The %d", n);        if (n % 10 == 1 && n % 100 != 11) printf("st ");        else if (n % 10 == 2 && n % 100 != 12) printf("nd ");        else if (n % 10 == 3 && n % 100 != 13) printf("rd ");        else printf("th ");        printf("humble number is %d.\n", dp[n]);    }    return 0;}