[ACM_ZJUT_1058]Humble Numbers

Humble Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8472 Accepted Submission(s): 3684
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
InputThe 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 1 2 3 4 11 12 13 21 22 23 100 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.
Source HDU1058
今日は授業が多くて、夜もあります.ちょうど水の問題を見て、数えてみましたが、悩んでいるのはWrong Answerです.その後、私の英語のレベルが向上しなければならないことに気づいた.11,12,13の後ろはthを加えただけでなく、11,12,13で終わった数字の後ろはst,nd,rdではなくthを加えた.
考え方は[ACM_ZJUT_1089]Ugly Numbersとほぼそっくりで、ダイナミックな計画で、あまり話さない.
コードは以下の通りです(時間が遅く、明日は朝の授業があるので、あまり最適化されていません.見苦しいかもしれません):

#include<stdio.h>
#include<vector>
#include<cmath>
using namespace std;
//_hdu_1058
int main(){
	int n = 5842, m;
	vector<int> v;
	int x2 = 0, x3 = 0, x5 = 0, x7 = 0;
	char s[4][3] = {"th", "st", "nd", "rd"};
	v.push_back(1);
	while(--n){
		int r2 = v[x2] * 2;
		int r3 = v[x3] * 3;
		int r5 = v[x5] * 5;
		int r7 = v[x7] * 7;
		m = min(r2, r3);
		m = min(m, r5);
		m = min(m, r7);
		v.push_back(m);
		if(r2 == m)
			++x2;
		if(r3 == m)
			++x3;
		if(r5 == m)
			++x5;
		if(r7 == m)
			++x7;
	}
	while(scanf("%d", &n) && n){
		printf("The %d%s humble number is %d.
", n, (n % 10 < 4 && n % 100 != 11 && n % 100 != 12 && n % 100 != 13)?s[n % 10]:"th", v[n - 1]);
	}
	return 0;
}

=================================================しかし、少なくとも私はそれのためにもっと努力することができます.署名ファイル===================