习题三1004
2016-05-06 17:58
381 查看
Problem D
TimeLimit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K(Java/Other)Problem Description
Anumber whose only prime factors are 2,3,5 or 7 is called a humble number. Thesequence 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. <br><br>Write a program tofind
and print the nth element in this sequence<br>
Input
Theinput consists of one or more test cases. Each test case consists of oneinteger n with 1 <= n <= 5842. Input is terminated by a value of zero (0)for n.<br>
Output
Foreach test case, print one line saying "The nth humble number isnumber.". Depending on the value of n, the correct suffix "st","nd", "rd", or "th" for the o
a6b6
rdinal number nthhas to be used like it is shown in the sample output.<br>
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:
符合要求的丑数只含有2、3、5、7的质因子求前5842个丑数。
解题思路:
humble number从1为"始祖",剩下的所有数,其实都是在此基础上乘以2,3,5,7演化出来的,:f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);
#include <iostream>
#include <stdio.h>
using namespace std;
int f[5843],n;
int i,j,k,l;
int min(int a,int b,int c,int d)
{
int min=a;
if(b<min) min=b;
if(c<min) min=c;
if(d<min) min=d;
if(a==min) i++;
if(b==min) j++;
if(c==min) k++;
if(d==min) l++;
return min;
}
int main()
{
i=j=k=l=1;
f[1]=1;
for(int t=2;t<=5842;t++)
{
f[t]=min(2*f[i],3*f[j],5*f[k],7*f[l]);
}
while(scanf("%d",&n)&&n!=0)
{
if(n%10==1&&n%100!=11)
printf("The %dst humble number is %d.\n",n,f
);
else if(n%10==2&&n%100!=12)
printf("The %dnd humble number is %d.\n",n,f
);
else if(n%10==3&&n%100!=13)
printf("The %drd humble number is %d.\n",n,f
);
else
printf("The %dth humble number is %d.\n",n,f
);
}
return 1;
}
相关文章推荐
- SQLServer分区表
- 纯干货!大前端必备网站-超全(上
- JS获取客户端IP地址、MAC和主机名的7个方法汇总
- 【krpano】二维码自动生成插件(源码+介绍+预览)
- Python 应用剖析工具介绍
- jQuery确定删除提示框
- Java访问控制权限
- Killing container
- Java访问控制权限
- Java访问控制权限
- 开发app需要角色
- linux环境变量
- Nginx教程(四) Location配置与ReWrite语法
- memcached搭建
- JavaScript之实现基本的增删改查功能
- EF小节
- 从零开始学_JavaScript_系列(17)——CSS<4>(定位、遮罩、float、弹性布局flex)
- android文件上传
- SAP 一些国企的状态
- Warning: The name 'layoutWidget' (QWidget) is already in use, defaulting to 'layoutWidget1'.