1069. The Black Hole of Numbers (20)

標籤：pat   演算法   stringstream   vector   

題目如下：

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 = 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

題目要求將一個四位元按照降序、升序產生減數和被減數然後作差，如果差等於0或者6174則停止運算。

題目的注意點在於數字不夠4位時要前補0。

為了方便的排序四位元，我們先使用vector容納四位然後排序，接著利用stringstream將vecotr中元素轉成數字，因為輸出時減數、被減數和差都需要，因此用一個結構體儲存每次的運算結果。

代碼如下：

#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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1069. The Black Hole of Numbers (20)