題目原文
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
題目大意
就是給你一個數(0-9999)然後讓你用這個數字 個位,十位,百位,千位上的4個數字,組成一個最大的數,和一個最小的數.
然後它們相減. 相減後的數字如果是6174或者0就停止.
不然就一直這樣繼續.
我的思路:
就是用一個數組保存這4個數字,然後算出最大和最小.
然後一直循環.
我的代碼:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int min = 0, max = 0,New;
vector<int> inex;
cin >> New;
do{
int t = 0;
while (New) {
t = New % 10;
inex.push_back(t);
New /= 10;
}
while (inex.size() != 4) {
inex.push_back(0);
}
sort(inex.begin(), inex.end());
min=0;
min=inex[0]*1000+inex[1]*100+inex[2]*10+inex[3];
max=inex[3]*1000+inex[2]*100+inex[1]*10+inex[0];
New = max-min;
printf("%04d - %04d = %04d\n",max,min,New);;
inex.clear();
}while(New!=0&&New%1111!=0&&New!=6174);
return 0;
}
柳神的代碼
#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(char a, char b) {return a > b;}
int main() {
string s;
cin >> s;
s.insert(0, 4 - s.length(), '0');
do {
string a = s, b = s;
sort(a.begin(), a.end(), cmp);
sort(b.begin(), b.end());
int result = stoi(a) - stoi(b);
s = to_string(result);
s.insert(0, 4 - s.length(), '0');
cout << a << " - " << b << " = " << s << endl;
} while (s != "6174" && s != "0000");
return 0;
}
柳神的代碼和我的差不多,不過比起我的代碼,她的代碼高度集成化.