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@xunuo 2016-08-13T17:17:06.000000Z 字数 1917 阅读 1515

Humble Numbers

丑数
HDU 1058
Time Limit:1000MS  Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

动态规划(DP)

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:

如果一个数a[i]只能被2,3,5,7整除,那么它就是丑数,现在让你找出第n个丑数是多少。
除11,12,13外,如果n%10==1则为st后缀,如果n%10==2,则为nd后缀,如果n%10==3,则为rd后缀,其余都为th后缀(包括n%100==11,12,13)。

完整代码:

  1. #include<stdio.h>
  2. #include<algorithm>
  3. using namespace std;
  4. int main()
  5. {
  6. int a[10000],i,m;
  7. a[0]=1;
  8. int x2=0;
  9. int x3=0;
  10. int x5=0;
  11. int x7=0;
  12. for(i=1;i<=5842;i++)
  13. {
  14. a[i]=min(min(min(a[x2]*2,a[x3]*3),a[x5]*5),a[x7]*7);
  15. if(a[i]==a[x2]*2)
  16. x2++;
  17. if(a[i]==a[x3]*3)
  18. x3++;
  19. if(a[i]==a[x5]*5)
  20. x5++;
  21. if(a[i]==a[x7]*7)
  22. x7++;
  23. }
  24. //将丑数全部存入后再找,,不然输入一个n,找一次,,要超时!!!!
  25. int n;
  26. while(scanf("%d",&n),n)
  27. {
  28. if(n%100==11||n%100==12||n%100==13)
  29. printf("The %dth humble number is %d.\n",n,a[n-1]);
  30. else if(n%10==1)
  31. printf("The %dst humble number is %d.\n",n,a[n-1]);
  32. else if(n%10==2)
  33. printf("The %dnd humble number is %d.\n",n,a[n-1]);
  34. else if(n%10==3)
  35. printf("The %drd humble number is %d.\n",n,a[n-1]);
  36. else
  37. printf("The %dth humble number is %d.\n",n,a[n-1]);
  38. }
  39. return 0;
  40. }

此题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1058
相似题目:https://www.oj.swust.edu.cn/problem/show/1026

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