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
解题思路:用string来做比较方便,排序的话直接用sort(s.begin(),s.end(),cmp),字符直接比较,我居然还把它转化成数字再来比较,后面就直接字符比较,然后计算差的时候直接把它们转成数字,这里有个注意的地方,这题还算有点坑,就是它输入的不一定是4位数,还要自己再insert‘0’。
#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(0,4-s.size(),'0');
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(0,4-s.size(),'0');
cout<<a<<" - "<<b<<" = "<<s<<endl;
}while(s!="6174"&&s!="0000");
return 0;
}