标签：tom int term otto one ott 就是 题意 amp
Humble Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30212 Accepted Submission(s): 13229
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
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：输入n，输出第n个丑数
题解：丑数是可以由2，3，5，7，因子组成的，应用丑数是2，3，5，7的倍数这一性质来做。用动态规划，从第一个丑数1开始，第二个丑数就是前面的丑数乘以2，3，5，7后最小的那个值，就是2。
1 #include<bits/stdc++.h>
2 using namespace std;
3 long long a[5845];
4 int main() {
5 int num=1;
6 int p2,p3,p5,p7;
7 p2=p3=p5=p7=1;
8 a[1]=1;
9 while(num<5843) {
10 a[++num]=min(min(a[p2]*2,a[p3]*3),min(a[p5]*5,a[p7]*7));
11 if(a[num]==2*a[p2])//找到之后就在下面用if语句把能得到当前
12 p2++; // 丑数的情况去掉 继续找下一个
13 if(a[num]==3*a[p3])
14 p3++;
15 if(a[num]==5*a[p5])
16 p5++;
17 if(a[num]==7*a[p7])
18 p7++;
19 }
20 int n;
21 while(~scanf("%d",&n),n)
22 {
23 if(n%10==1&&n%100!=11)
24 printf("The %dst humble number is %lld.\n",n,a[n]);
25 else if(n%10==2&&n%100!=12)
26 printf("The %dnd humble number is %lld.\n",n,a[n]);
27 else if(n%10==3&&n%100!=13)
28 printf("The %drd humble number is %lld.\n",n,a[n]);
29 else
30 printf("The %dth humble number is %lld.\n",n,a[n]);
31 }
32
33 return 0;
34 }
hdu1058Humble Numbers(动态规划)
标签：tom int term otto one ott 就是 题意 amp
原文地址：https://www.cnblogs.com/fqfzs/p/9842476.html
