PAT 甲级 1069 The Black Hole of Numbers (20 分)(内含别人string处理的精简代码)
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 (.
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
题意:
给一个数组n,先对各个数字位从大到小排序得到a,然后逆置得到b,计算a-b,得到c,重复上述过程,直到得到6174,或者,对于特例,四个数字位完全相同,那么输出0000。
题解:
刚开始输入的数字可能不足四位要自动补0 以后的结果可能也不足四位都要自动补0
结果是0或者6174都是要退出的
一开始没考虑到6174,测试点5没过,后来想到了
AC代码:
#include<iostream>
#include<algorithm>
using namespace std;
int n;
int a[];
int main(){
cin>>n;
int big=;
int small=;
int x=n;
int k=,f;
if(x==){
printf("7641 - 1467 = 6174");
return ;
}
while(x!=){
k=;
big=;
small=;
while(x>=){
a[++k]=x%;
x/=;
}
a[++k]=x;
while(k<){
a[++k]=;
}
sort(a+,a++);
for(int i=;i<=;i++){
big=big*+a[-i];
small=small*+a[i];
}
x=big-small;
printf("%04d - %04d = %04d\n",big,small,x);
if(x==) break;
}
return ;
}
更简洁的string处理的代码:
#include<bits/stdc++.h>
using namespace std;
bool cmp(char a,char b)
{
return a>b;
}
int main(void)
{
string s;
cin>>s;
s.insert(,-s.size(),'');
do{
string a=s,b=s;
sort(a.begin(),a.end(),cmp);
sort(b.begin(),b.end());
int diff=stoi(a)-stoi(b);
s=to_string(diff);
s.insert(,-s.size(),'');
cout<<a<<" - "<<b<<" = "<<s<<endl;
}while(s!=""&&s!="");
return ; ————————————————
版权声明:本文为CSDN博主「Imagirl1」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/Imagirl1/article/details/82261213
PAT 甲级 1069 The Black Hole of Numbers (20 分)(内含别人string处理的精简代码)的更多相关文章
- 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 ...
- 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甲级】1023 Have Fun with Numbers (20 分)
题意: 输入一个不超过20位的正整数,问乘2以后是否和之前的数组排列相同(数字种类和出现的个数不变),输出Yes或No,并输出乘2后的数字. AAAAAccepted code: #define HA ...
- 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:1124 Raffle for Weibo Followers (20分)
PAT甲级:1124 Raffle for Weibo Followers (20分) 题干 John got a full mark on PAT. He was so happy that he ...
- 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 ...
- 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部署TOMCAT服务器
转载声明: http://www.cnblogs.com/xdp-gacl/p/4097608.html 解压tomcat服务器压缩包 配置环境变量 tomcat服务器运行时是需要JDK支持的,所以必 ...
- springcloud实践(二)之api网关:zuul
zuul是什么? front door. API Gateway.Zuul is a JVM based router and server side load balancer by Netflix ...
- JAVA读取XML并打印
在G盘下新建XML文档:person.xml,XML代码: <?xml version="1.0" encoding="utf-8"?> <s ...
- 配置Sublime,为了Python
E:\Sublime Text 3\Data\Packages\User\untitled.sublime-build { "cmd": ["C:\Program Fil ...
- 洛谷 P1194 买礼物 题解
P1194 买礼物 题目描述 又到了一年一度的明明生日了,明明想要买\(B\)样东西,巧的是,这\(B\)样东西价格都是\(A\)元. 但是,商店老板说最近有促销活动,也就是: 如果你买了第II样东西 ...
- github上项目的目录结构说明
build 构建脚本 dist 编译出来的发布版 docs 文档 examples 示例文件 src 源码 test 测试脚本 .babelrc Babel 交叉编译的配置 .eslintrc ESL ...
- (2)Go基本数据类型
Go语言的基本类型有: bool string int.int8.int16.int32.int64 uint.uint8.uint16.uint32.uint64.uintptr byte // u ...
- UOJ#318. 【NOI2017】蔬菜 贪心
原文链接 www.cnblogs.com/zhouzhendong/p/UOJ318.html 前言 我怎么越来越菜了.先是题目读错,想了个李超树假算法,然后读懂题之后没了耐心直接贺题.然后发现我数据 ...
- os 模块常用方法
os.remove()删除文件 os.rename()重命名文件 os.walk()生成目录树下的所有文件名 os.chdir()改变目录 os.mkdir/makedirs创建目录/多层目录 os. ...
- Redis哨兵日常实践
一.日常操作 指定一个从做新主 有时候需要将当前主节点机器下线,并指定一个高一些性能的从节点接替 将其它从节点的slave-priority配置为0,然后在随意一台 Setinel 执行sentine ...