HDU 1058 Humble Numbers
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
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:
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.
题意:因子只有2,3,5,7的数被称为humble number,1是例外,那么第n个这样的数是几呢,注意这边的输出也有坑。。。。
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int N=6000; int a[N]; void Humble() { int x2, x3, x5, x7, y2=1, y3=1, y5=1, y7=1, i; a[1] = 1; for (i = 2; i <= 5842; i++) { x2 = a[y2]*2; x3 = a[y3]*3; x5 = a[y5]*5; x7 = a[y7]*7; a[i] = min(x2, x3); a[i] = min(a[i], x5); a[i] = min(a[i], x7); if (x2 == a[i]) y2++; ///x2,x3,x4,x5的最小值作为a[i],那么下标要加1,以免重复 if (x3 == a[i]) y3++; if (x5 == a[i]) y5++; if (x7 == a[i]) y7++; } } int main () { int n; Humble(); while (scanf("%d", &n), n) { if (n % 10 == 1 && n/10%10 != 1) printf("The %dst humble number is %d.\n", n, a[n]); ///注意英语中第几个数的输出,个位为1,2,3,但是十位不能是1,2,3,才能输出st,nd,rd,其他输出th else if (n % 10 == 2 && n/10%10 != 1) printf("The %dnd humble number is %d.\n", n, a[n]); else if (n % 10 == 3 && n/10%10 != 1) printf("The %drd humble number is %d.\n", n, a[n]); else printf("The %dth humble number is %d.\n", n, a[n]); } return 0; }