PAT 1069. The Black Hole of Numbers (20)

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

判题程序

Standard

作者

CHEN, Yue

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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

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这道题我还是把它当成字符串处理问题做的。 代码如下：

#include <iostream>#include <algorithm>#include <cmath>#include <string>#include <iomanip>#include <cstring>using namespace std;int main(void){string raw;cin>>raw;int i;int temp=4-raw.size();if(temp>0){string t="";for(i=0;i<temp;i++)t+='0';t+=raw;raw=t;}for(i=0;i<raw.size();i++)if(raw[i]!=raw[0])break;if(i>=raw.size()){cout<<raw<<" - "<<raw<<" = "<<"0000";return 0;}sort(raw.begin(),raw.end());string rev=raw;reverse(rev.begin(),rev.end());int result=atoi(rev.c_str())-atoi(raw.c_str());while(result!=6174){cout<<rev<<" - "<<raw<<" = ";char buf[5];sprintf(buf,"%d",result);rev=buf;int j=4-rev.size();raw="";for(i=0;i<j;i++)raw+='0';raw+=rev;cout<<raw<<endl;sort(raw.begin(),raw.end());rev=raw;reverse(rev.begin(),rev.end());result=atoi(rev.c_str())-atoi(raw.c_str());}cout<<rev<<" - "<<raw<<" = "<<result;}