PAT (Advanced Level) Practise 1069 The Black Hole of Numbers (20)

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本文介绍了一种数学游戏，通过不断重组四位数并求差，最终达到固定点6174，即Kaprekar常数。文章提供了一个C++程序示例，用于演示这一过程。

1069. The Black Hole of Numbers (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

Sample Output 1:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174

Sample Input 2:

Sample Output 2:

2222 - 2222 = 0000

题意：给你一个数，将它所有数字从大到小排序形成一个数，所有数字从小到大排序形成一个数，然后相减，形成一个形的数，不断地这样操作，直至它为0或为6174

解题思路：暴力模拟

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, a[20];

int get(int x, int flag)
{
	for (int i = 0; i < 4; i++) a[i] = x % 10, x /= 10;
	if (flag) sort(a, a + 4);
	else sort(a, a + 4, greater<int>());
	int ans = 0;
	for (int i = 0; i < 4; i++) ans = ans * 10 + a[i];
	return ans;
}

int main()
{
	while (~scanf("%d", &n))
	{
		do
		{
			int x = get(n, 0), y = get(n, 1);
			n = x - y;
			printf("%04d - %04d = %04d\n", x, y, n);
		}while (n != 6174 && n);
	}
	return 0;
}