PAT 1069. The Black Hole of Numbers (20)（数字黑洞）

官网

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:
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

解题思路

• 1.把输入数字的每个位上的数字都提取出来，然后排序再相减，看结果是否为0或者为6174，如果都不是则继续循环。
• 2.注意输入为1，11，111这种情况。

AC代码

#include<iostream>#include<string>#include<vector>#include<math.h>#include<algorithm>#include<cstdio>using namespace std;int main(int argc, char *argv[]){    int a,b = 0,c = 0;    cin >> a;    if (a == 0) {        printf("0000 - 0000 = 0000\n");        return 0;    }    do {        b = 0,c = 0;        vector<int> keep;        while (a>0) {            keep.push_back(a%10);            a = a / 10;        }        //排序求最大值        sort(keep.begin(),keep.end(),greater<int>());        int len = keep.size();        while (len<4) {            keep.push_back(0);            len++;        }        for (int i = 0; i < len; ++i) {            b += keep[i] * pow(10,len - i - 1);        }        //排序求最小值        sort(keep.begin(),keep.end(),less<int>());        len = keep.size();        while (len<4) {            keep.push_back(0);            len++;        }        for (int i = 0; i < len; ++i) {            c += keep[i] * pow(10,len - i - 1);        }        a = b - c;        printf("%04d - %04d = %04d\n",b,c,a);    } while (a!=6174 && a != 0);    return 0;}