PAT 1069 The Black Hole of Numbers[简单]
1069 The Black Hole of Numbers(20 分)
For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner we will soon end up at the number 6174 -- the black hole of 4-digit numbers. This number is named Kaprekar Constant.
For example, start from 6767, we'll get:
7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174
7641 - 1467 = 6174
... ...
Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.
Input Specification:
Each input file contains one test case which gives a positive integer N in the range (0,104).
Output Specification:
If all the 4 digits of N are the same, print in one line the equation N - N = 0000. Else print each step of calculation in a line until 6174 comes out as the difference. All the numbers must be printed as 4-digit numbers.
Sample Input 1:
6767
Sample Output 1:
7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174
Sample Input 2:
2222
Sample Output 2:
2222 - 2222 = 0000
题目大意:给定一个四位数,展示出每位数从大到小排列,与从小到大排列的差值,直到出现6174黑洞数停止。
//这道题目是简单的,但是还是遇见了一些问题:
AC:
#include <iostream>
#include <algorithm>
#include <vector>
#include <stdlib.h>
#include<cstdio>
using namespace std; int main()
{
int n;
cin>>n;
int a[],big,small,res=-;
int temp=;
while(res!=)
{
fill(a,a+,);
while(n!=)
{
a[temp++]=n%;
n/=;
//cout<<"kk";
}
temp=;
sort(a,a+);//默认从小到大排列
small=a[]*+a[]*+a[]*+a[];
big=a[]*+a[]*+a[]*+a[];
if(a[]==a[]&&a[]==a[]&&a[]==a[])
{
printf("%04d - %04d = 0000\n",big,small);
return ;
}
res=big-small;
printf("%04d - %04d = %04d\n",big,small,res);
n=res;
}
return ;
}
1.第一次忽略了temp=0;应该在使用过后将其赋值为0的;
2. 应该将a数组初始化为0,要不然下次会有影响的,比如在有0位的时候:
3.最后就是提交发现第0个测试点过不去,原来是因为相同数的时候只输出了0,而不是0000!
PAT 1069 The Black Hole of Numbers[简单]的更多相关文章
- PAT 1069 The Black Hole of Numbers
1069 The Black Hole of Numbers (20 分) For any 4-digit integer except the ones with all the digits ...
- pat 1069 The Black Hole of Numbers(20 分)
1069 The Black Hole of Numbers(20 分) For any 4-digit integer except the ones with all the digits bei ...
- PAT 1069. The Black Hole of Numbers (20)
For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in ...
- 1069. The Black Hole of Numbers (20)【模拟】——PAT (Advanced Level) Practise
题目信息 1069. The Black Hole of Numbers (20) 时间限制100 ms 内存限制65536 kB 代码长度限制16000 B For any 4-digit inte ...
- PAT 甲级 1069 The Black Hole of Numbers (20 分)(内含别人string处理的精简代码)
1069 The Black Hole of Numbers (20 分) For any 4-digit integer except the ones with all the digits ...
- 1069 The Black Hole of Numbers (20分)
1069 The Black Hole of Numbers (20分) 1. 题目 2. 思路 把输入的数字作为字符串,调用排序算法,求最大最小 3. 注意点 输入的数字的范围是(0, 104), ...
- PAT Advanced 1069 The Black Hole of Numbers (20) [数学问题-简单数学]
题目 For any 4-digit integer except the ones with all the digits being the same, if we sort the digits ...
- PAT (Advanced Level) 1069. The Black Hole of Numbers (20)
简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...
- PAT甲题题解-1069. The Black Hole of Numbers (20)-模拟
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789244.html特别不喜欢那些随便转载别人的原创文章又不给 ...
随机推荐
- Linux - 用户管理常用命令
1.查看Linux已经存在的用户: [root@CMCC_91 ~]# cut -d : -f 1 /etc/passwd [root@CMCC_91 ~]# cat /etc/passwd |awk ...
- JAVASCRIPT+DHTML实现表格拖动
自已做的,本来想在网上找前辈们做的,可是总找不到这种例子,要么找出来的太复杂, 要么就没法用,索性自己写了一个.看看还可以用!贡献出来,估计和我一样的菜鸟用的着! <html> <s ...
- 22SpringMvc_jsp页面上的数据传递到控制器的说明
假设有这个一个业务:在jsp页面上写入数据,然后把这个数据传递到后台. 效果如下:
- "reason":"No handler for type [attachment] declared on field [file]" 最完全解决方案
0.elasticsearch-mapper-attachments 2.3.4安装 mapper-attachments安装方法分两类,在线和离线: 在线安装 bin/elasticsearch-p ...
- WORD里怎样能做到局部“分栏”就是一页里有的分有的不分
选中你要分的部分再分栏如果不想分的部分也被分了,那就可以选中不想分的那部分,选择“分栏”->“一栏” 转自:http://zhidao.baidu.com/question/9873268.ht ...
- JSP 页面中用绝对路径显示图片
首先,图片和工程不在一个盘符下.图片也不能放到工程下. 在JSP 文件中 <img src="E:/图片/1.jpg"/> 这样是引不到图片的.因为,JSP页面在引 ...
- Maven仓库的搭建
http://blog.csdn.net/xiao__gui/article/details/52625660 Maven仓库是有特定规则的目录结构. 目录结构由 仓库根目录 , groupId , ...
- PE导入表分析
A.dll 导入 B.dll 导出函数 A.dll 表内容 这个结构指向的B导出函数的地址 Hook这个位置 等同于 Hook B.dll导出函数
- android基础---->AccessibilityService的简单使用(一)
AccessibilityService类可以帮助我们实现监听手机上别的应用,以下做一个简单的总结.我总是勇敢的离开一个人 却不懂如何巧妙的靠近一个人. AccessibilityService的使用 ...
- Javascript中的window.event.keyCode使用介绍
<body onkeydown=" alert(window.event.keyCode)"> <body onkeydown="if(window.e ...