hdu1058 Humble NUMBERS--DP

Original title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1058

One: The original question content

Problem Description A number whose only prime factors are 2,3,5/7 is called a humble number. The sequence is 1, 2, 3, 4, 5, 6, 7, 8, 9, and, A,,, and, shows the numbers of the "humble".

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. The Input is terminated by a value of zero (0) for N.
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 is used like I T is shown in the sample output.
Sample Input

1 2 3 4 1000 5842 0 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. Second: Analysis and understanding

The whole idea is to use the elements in the known sequence to generate new elements according to the rules, such as X * 2, X * 3, X * 5, X * 7.
Suppose you use array A[max] to store all the elements of this sequence, first make a[0] = 1, which means that the first element is 1, then use 4 pointers (not memory pointers, hehe), i2 = i3 = i5 = i7 = 0. Each time we compare A[I2] * 2, A[I3] * 3, A[I5] * 5, A[i7] * 7 The size of the four elements, put the smallest into the sequence, and put the corresponding pointer +1 until the number of required numbers is generated.

Three: AC code

#include <iostream>  
#include <string.h>
#include <algorithm>  

using namespace std;

int a[6000];

int main ()
{
	a[1] = 1;
	int I2, i3, i5, i7;
	I2 = i3 = i5 = i7 = 1;

	for (int i = 2; I <= 5842 i++)
	{
		int x = min (min (a[i2) * 2, A[i3] * 3), MIN (a[i5] * 5, A[i7] * 7));

		if (x = = A[i2] * 2) i2++;//This place cannot use the If...else statement, because the comparison of four numbers may exist equal
		if (x = = A[i3] * 3) i3++;
		if (x = = a[i5] * 5) i5++;
		if (x = = A[i7] * 7) i7++;

		A[i] = x;
	}

	int n;

	while (scanf ("%d", &n) && N)
	{
		if (n% = 1 && n%!=)
			printf ("The%DST HUMBL E number is ", n);
		else if (n% = 2 && n%!=)
			printf ("The%DND humble number is", n);
		else if (n% = 3 && n%!=)
			printf ("The%DRD humble number is", n);
		else
			printf ("The%dth humble number is", n);

		printf ("%d.\n", A[n]);
	
	return 0;
}