1069. The Black Hole of Numbers (20)

分类：算法学习

题目如下：

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 = 10899810 – 0189 = 96219621 – 1269 = 83528532 – 2358 = 61747641 – 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 1:6767Sample Output 1:7766 – 6677 = 10899810 – 0189 = 96219621 – 1269 = 83528532 – 2358 = 6174Sample Input 2:2222Sample Output 2:2222 – 2222 = 0000

题目要求将一个四位数按照降序、升序生成减数和被减数然后作差，如果差等于0或者6174则停止运算。

题目的注意点在于数字不够4位时要前补0。

为了方便的排序四位数，我们先使用vector容纳四位然后排序，接着利用stringstream将vecotr中元素转成数字，因为输出时减数、被减数和差都需要，因此用一个结构体存储每次的运算结果。

代码如下：

#include <iostream>#include <vector>#include <sstream>#include <algorithm>#include <sstream>#include <stdio.h>using namespace std;struct Result{int num1;int num2;int result;}res;void compute(int input){vector<int> num1,num2;num1.push_back(input / 1000);num1.push_back(input % 1000 / 100);num1.push_back(input % 100 / 10);num1.push_back(input % 10);num2 = num1;sort(num1.begin(),num1.end(),less_equal<int>());sort(num2.begin(),num2.end(),greater_equal<int>());stringstream ss;for(int i = 0; i < 4; i++){ss << num1[i];}int dec1;ss >> dec1;ss.clear();for(int i = 0; i < 4; i++){ss << num2[i];}int dec2;ss >> dec2;res.num1 = dec2;res.num2 = dec1;res.result = dec2 – dec1;}int main(){int input;cin >> input;int i = 5;do{compute(input);input = res.result;if(input == 0){printf("%04d – %04d = %04d\n",res.num1,res.num2,input);break;}else{printf("%04d – %04d = %04d\n",res.num1,res.num2,input);}}while(input != 6174);return 0;}

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