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 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, 10000).
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
解题思路
直接计算就可以
AC代码
#include <cstdio>
#include <algorithm>
using namespace std;
int getmin(int a){
int t[4] = {0}, c = 0;
while (a){
t[c++] = a%10;
a /= 10;
}
sort(t, t + 4);
c = 0;
while (c < 4){
a = a*10 + t[c++];
}
return a;
}
int getmax(int a){
int t[4] = {0}, c = 0;
while (a){
t[c++] = a%10;
a /= 10;
}
sort(t, t + 4, greater<int>());
c = 0;
while (c < 4){
a = a*10 + t[c++];
}
return a;
}
int main()
{
int a;
scanf("%d", &a);
while (true){
int mn = getmin(a);
int mx = getmax(a);
printf("%04d - %04d = %04d\n", mx, mn, mx - mn);
a = mx - mn;
if (a == 6174 || mn%10 == mn/1000) break;
}
return 0;
}
个人游戏推广:
《10云方》与方块来次消除大战!
1069. The Black Hole of Numbers (20)【模拟】——PAT (Advanced Level) Practise的更多相关文章
- 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 ...
- 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 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)
For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in ...
- 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甲级】1069 The Black Hole of Numbers (20 分)
题意: 输入一个四位的正整数N,输出每位数字降序排序后的四位数字减去升序排序后的四位数字等于的四位数字,如果数字全部相同或者结果为6174(黑洞循环数字)则停止. trick: 这道题一反常态的输入的 ...
- 1084. Broken Keyboard (20)【字符串操作】——PAT (Advanced Level) Practise
题目信息 1084. Broken Keyboard (20) 时间限制200 ms 内存限制65536 kB 代码长度限制16000 B On a broken keyboard, some of ...
随机推荐
- COGS——T 1786. 韩信点兵
http://www.cogs.pro/cogs/problem/problem.php?pid=1786 ★★★ 输入文件:HanXin.in 输出文件:HanXin.out 简单对比时 ...
- C# 反射具体解释
watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/ ...
- 如何让Apache不显示服务器信息
如何让Apache不显示服务器信息 Apache的默认配置是会显示服务器信息的,比如访问一个服务器上不存在的页面,Apache会返回"Not Found"的错误,这个错误页面的最下 ...
- noi25 最长最短单词(为什么会出现运行时错误)
noi25 最长最短单词(为什么会出现运行时错误) 一.总结 一句话总结:比如除以零,数组越界,指针越界,使用已经释放的空间,数组开得太大,超出了栈的范围,造成栈溢出 1.c++报runtime er ...
- 关于python的拷贝
https://blog.csdn.net/koukehui0292/article/details/82993958 Python的 深度拷贝: import copy d=copy.deepcop ...
- python3 turtle 画围棋棋盘
python3 环境 利用turtle模块画出 围棋棋盘 #!/usr/bin/env python # -*- coding:utf-8 -*- # Author:Hiuhung Wan impor ...
- 发布一个stl标准库容器类(vector/list)的安全删除方法
话不多说,看代码. #include <functional> #ifndef ASSERT #include <cassert> #define ASSERT assert ...
- 读<阿里亿级日活网关通道架构演进>有感
读<阿里亿级日活网关通道架构演进>时对优化方法有些概念不理解,特意搜索了一下,拓展自己的思路. 其中的优化: 优化方法中1,2比较常见,3,4我知道的比较少,很感兴趣.就继续追踪下去: 于 ...
- 【AtCoder ABC 075 C】Bridge
[链接] 我是链接,点我呀:) [题意] 让你求出桥的个数 [题解] 删掉这条边,然后看看1能不能到达其他所有的点就可以了 [代码] #include <bits/stdc++.h> us ...
- 【Codeforces Round #434 (Div. 2) A】k-rounding
[链接]h在这里写链接 [题意] 在这里写题意 [题解] 转换一下就是求n和10^k的最小公倍数. [错的次数] 0 [反思] 在这了写反思 [代码] #include <bits/stdc++ ...