1069. The Black Hole of Numbers(20) - PAT甲级

原创 于 2019-07-11 16:40:09 发布 · 296 阅读 · 0 ·
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博客围绕PAT中4位数黑洞6174的题目展开，介绍了题目内容，即对非全相同的4位数排序相减最终会得到6174。作者分享了自己的AC代码编写过程，提到遇到的两个坑，一是要注意结果为四位数，二是pow函数精度问题，最后还学习了柳神的代码及C++ STL相关知识。

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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, 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 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174
Sample Input 2:
2222
Sample Output 2:
2222 - 2222 = 0000

我的AC代码（刚开始刷PAT，很菜）：

#include <iostream>
#include <string>
#include <math.h>
using namespace std;

int sort(int x) {
	int t[4];
	for (int i = 0; i < 4; i++) {
		t[i] = x % (int)round(pow(10, i + 1)) / (int)round(pow(10, i));
        x = x - t[i]* (int)round(pow(10, i));
	}
	int temp;
	int m, n;
	for (m = 0; m < 3; m++) {
		for (n = 0; n < 3 - m; n++) {
			if (t[n] < t[n + 1]) {
				temp = t[n];
				t[n] = t[n + 1];
				t[n + 1] = temp;
			}
		}
	}
	return t[0] * 1000 + t[1] * 100 + t[2] * 10 + t[3];
}

int nixu(int x) {
	int t[4];
	for (int i = 0; i < 4; i++) {
		t[i] = x % (int)round(pow(10, i + 1)) / (int)round(pow(10, i));
	}
	return t[0] * 1000 + t[1] * 100 + t[2] * 10 + t[3];
}
int main() {
	int input;
	cin >> input;
	int num1, num2, res;
	num1 = sort(input);
	num2 = nixu(num1);
	res = num1 - num2;
	if(num1 == 0){				//当num1=0时对num1前面加0，不然输出为0-0000=0
		if(num2 < 1000 && num2 >= 100){
			cout << "000" << num1 << " - 0" << num2 << " = 000" << res << endl;
		}
		else if(num2 < 100 && num2 >= 10){
			cout << "000" << num1 << " - 00" << num2 << " = 000" << res << endl;
		}
		else if(num2 < 10 && num2 >= 0){
			cout << "000" << num1 << " - 000" << num2 << " = 000" << res << endl;
			}
		else{
			cout << "000" << num1 << " - " << num2 << " = 000" << res << endl;
		}
	}

	else if (num1 != 0 && res <= 9 & res >= 0){				//当num1!=0，对res加0，不然存在输出res不是四位数的情况
		if(num2 < 1000 && num2 >= 100){
			cout << num1 << " - 0" << num2 << " = 000" << res << endl;
		}
		else if(num2 < 100 && num2 >= 10){
			cout << num1 << " - 00" << num2 << " = 000" << res << endl;
		}
		else if(num2 < 10 && num2 >= 0){
				cout << num1 << " - 000" << num2 << " = 000" << res << endl;
			}
		else{
			cout << num1 << " - " << num2 << " = 000" << res << endl;
		}
	}

	else if (num1 != 0 && res <= 99 & res > 9){
		if(num2 < 1000 && num2 >= 100){
			cout << num1 << " - 0" << num2 << " = 00" << res << endl;
		}
		else if(num2 < 100 && num2 >= 10){
			cout << num1 << " - 00" << num2 << " = 00" << res << endl;
		}
		else if(num2 < 10 && num2 >= 0){
				cout << num1 << " - 000" << num2 << " = 00" << res << endl;
			}
		else{
			cout << num1 << " - " << num2 << " = 00" << res << endl;
		}
	}

	else if (num1 != 0 && res <= 999 & res > 99){
		if(num2 < 1000 && num2 >= 100){
			cout << num1 << " - 0" << num2 << " = 0" << res << endl;
		}
		else if(num2 < 100 && num2 >= 10){
			cout << num1 << " - 00" << num2 << " = 0" << res << endl;
		}
		else if(num2 < 10 && num2 >= 0){
				cout << num1 << " - 000" << num2 << " = 0" << res << endl;
			}
		else{
			cout << num1 << " - " << num2 << " = 0" << res << endl;
		}
	}

	else if (num1 != 0 && res > 999){
		if(num2 < 1000 && num2 >= 100){
			cout << num1 << " - 0" << num2 << " = " << res << endl;
		}
		else if(num2 < 100 && num2 >= 10){
			cout << num1 << " - 00" << num2 << " = " << res << endl;
		}
		else if(num2 < 10 && num2 >= 0){
				cout << num1 << " - 000" << num2 << " = " << res << endl;
			}
		else{
			cout << num1 << " - " << num2 << " = " << res << endl;
		}
	}

	while (num1 != num2 && res != 6174){
		num1 = sort(res);
		num2 = nixu(num1);
		res = num1 - num2;
		if(res > 999){				//当num1=0时对num1前面加0，不然输出为0-0000=0
			if(num2 < 1000 && num2 >= 100){
				cout << num1 << " - 0" << num2 << " = " << res << endl;
			}
			else if(num2 < 100 && num2 >= 10){
				cout << num1 << " - 00" << num2 << " = " << res << endl;
			}
			else if(num2 < 10 && num2 >= 0){
				cout << num1 << " - 000" << num2 << " = " << res << endl;
				}
			else{
				cout << num1 << " - " << num2 << " = " << res << endl;
			}
		}
		else{
			if(num2 < 1000 && num2 >= 100){
				cout << num1 << " - 0" << num2 << " = 0" << res << endl;
			}
			else if(num2 < 100 && num2 >= 10){
				cout << num1 << " - 00" << num2 << " = 0" << res << endl;
			}
			else if(num2 < 10 && num2 >= 0){
				cout << num1 << " - 000" << num2 << " = 0" << res << endl;
			}
			else{
				cout << num1 << " - " << num2 << " = 0" << res << endl;
			}
		}
	} 
	return 0;
}

刚开始只看英文题目和输入样例，只想到了最基本的将输入数字按位取出，再进行排序和逆序处理，踩了个坑，完全没注意题目要求的四位数，一直卡在这。想了好久，不停输入测试才想到num1、num2和res都要求是四位数。

这里我是这样对输入的数据进行取出每一位的

for (int i = 0; i < 4; i++) {
		t[i] = x % (int)round(pow(10, i + 1)) / (int)round(pow(10, i));
        x = x - t[i]* (int)round(pow(10, i));
	}

还踩了个坑，用的VSCODE，同样的输入1798，在线编译器输出结果与我的预期结果一致,sort之后都是9871，而在VSCODE上一直都是8710，怎么想都觉得我写的逻辑没错。我原来的sort()里是这样的

for (int i = 0; i < 4; i++) {
		t[i] = x % (int)pow(10, i + 1) / pow(10, i);
        x = x - t[i]* (pow(10, i);
	}

后面所有的pow()前面都加上了int强转还是没用，最后百度到，pow(10,2)是99.999999，(int)pow(10,2)之后是99……两种解决办法：1.自己写pow()函数，2.用round()四舍五入。于是用了round()最后才得救。

最终AC之后，觉得自己的代码也太低端了…于是去找了柳神（软院考研狗的偶像）的代码学习

#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');								//用来给不足4位的时候前面补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;
}

C++ STL:
sort(a.begin(), a.end(), cmp)：将字符串a按cmp的规则对a进行排序处理
stoi()：将数值类型的字符串转换为int类型。atoi()功能与stoi()相似，主要区别在于atoi()的参数是const char*，因此对于字符串str必须调用c_str()把这个string转换成const char类型的，而stoi()的参数类型是const string，不需要转化

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