PAT甲级 1069 字符串处理

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1069 The Black Hole of Numbers

分数 20

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 (0,104).

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

代码长度限制

16 KB

时间限制

200 ms

内存限制

64 MB

C++ (g++)

#include <bits/stdc++.h>
using namespace std;bool cmp1(char a, char b) {return a > b;
}bool cmp2(char a, char b) {return a < b;
}//高精度减法
//保证a>b;
string diff(string a, string b) {vector<int> A, B, C;for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');for (int i = b.size() - 1; i >= 0; i--) B.push_back(b[i] - '0');int t = 0;for (int i = 0; i < A.size(); i++) {t += A[i];if (i < B.size()) t -= B[i];C.push_back((t + 10) % 10);if (t < 0) t = -1;else t = 0;}string c;//if(C.size()>1&&C.back()==0) C.pop_back(); for (int i = C.size() - 1; i >= 0; i--) c += to_string(C[i]);return c;
}int main() {char n[5]={'0','0','0','0'};int k;scanf("%d",&k);string bb = to_string(k);int i=3,j=bb.size()-1;while(j>=0&&i>=0){n[i--]=bb[j--];}string last = "";int l = strlen(n);while (1) {char m1[5], m2[5];strcpy(m1, n);strcpy(m2, n);sort(m1, m1 + l, cmp1);sort(m2, m2 + l, cmp2);string a1 = m1;string a2 = m2;string s = diff(a1, a2);if (s == last) break;cout << a1 << " - " << a2 << " = " << s << endl;last = s;strcpy(n, s.c_str());}return 0;
}

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