1. 6174猜想 ,1955年,卡普耶卡(D.R.Kaprekar)研究了对四位数的一种变换:任给出四位数k0,用它的四个数字由大到小重新排列成一个四位数m,再减去它的反序数rev(m),得出数k1=m-rev(m),然后,继续对k1重复上述变换,得数k2.如此进行下去,卡普耶卡发现,无论k0是多大的四位数, 只要四个数字不全相同,最多进行7次上述变换,就会出现四位数6174.
2.需要注意输入的数字不一定是4位的,需要转化为4位的string进行处理,如输入1,2,3,4。
3.输入为6174时,应该输出7641 – 1467 = 6174。一开始卡在这个测试点了。
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
AC代码:
//#include<string> //#include <iomanip> //#include<stack> //#include<unordered_set> //#include <sstream> //#include "func.h" //#include <list> #include<unordered_map> #include<set> #include<queue> #include<map> #include<vector> #include <algorithm> #include<stdio.h> #include<iostream> #include<string> #include<memory.h> #include<limits.h> #include<stack> using namespace std; /* 测试案例(不一定是4位数): 6174 需要输出7641 - 1467 = 6174 1 2 3 4 5 6 */ bool cmp(const char&a, const char&b) { return a > b; } string num2str(int a) { a += 10000;//如果有0,则相当于补充千位,百位的0 string ans = ""; for (int i = 1; i < 5; i++) { char c = a % 10 + '0'; ans = c + ans; a /= 10; } return ans; } int str2num(string s) { return (s[0] - '0') * 1000 + (s[1] - '0') * 100 + (s[2] - '0') * 10 + (s[3] - '0'); } int main(void) { int num; cin >> num; string s = num2str(num); char c = s[0]; bool isSame = true; for (int i = 0; i < 4; i++) { if(s[i] != c) { isSame = false; break; } } if (isSame) {//如果所有位相同,直接输出0000 cout << s << " - " << s << " = 0000" << endl; } else { string ans = "";//这样处理使得输入为6174时,也能输出7641 - 1467 = 6174 while (ans != "6174") { sort(s.begin(), s.end(), cmp); string a = s;//大到小排列 sort(s.begin(), s.end()); string b = s;//小到大排列 int tmp = str2num(a) - str2num(b); ans = num2str(tmp); cout << a << " - " << b << " = " << ans << endl; s = ans; } } return 0; }