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PAT 1069. The Black Hole of Numbers (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, 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
此题没有什么难度,基本上就是两个可逆的转换:将一串数字(或者一个字符串)转换为一个整数,或者相反,而这两个转换都是很常见的,司空见惯了。对于此题值得注意的是,不要用字符串来处理(用诸如string、atoi,itoa【gcc上好像没有,可以用memset和sprintf代替】),会超时的!!!什么都不说了,按部就班就好了,请看代码:
#include <cstdio> #include <algorithm> #include <functional> #include <vector> using namespace std; const int blackHole=6174; const int digits=4; vector<int> int2vec(int n) { vector<int> buf(digits,0); for(int i=0;i<digits;++i,n/=10) { buf[i]=n%10; } return buf; } int vec2int(vector<int>& vec) { int n=0; int radix=1; for(int i=digits-1;i>=0;--i) { n+=radix*vec[i]; radix*=10; } return n; } bool beingTheSame(vector<int>& vec) { size_t size=vec.size(); for(int i=1;i<size;++i) { if(vec[0]!=vec[i]) return false; } return true; } int repeat(int n) { vector<int> vec=int2vec(n); sort(vec.begin(),vec.end(),greater<int>()); int first=vec2int(vec); sort(vec.begin(),vec.end()); int second=vec2int(vec); int difference=first-second; printf("%.4d - %.4d = %.4d\n",first,second,difference); return difference; } int _tmain(int argc, _TCHAR* argv[]) { freopen("1069.txt","r",stdin); int n; scanf("%d",&n); vector<int> vec=int2vec(n); if(beingTheSame(vec)) { printf("%.4d - %.4d = 0000\n",n,n); return 0; } n=repeat(n); while(blackHole!=n) { n=repeat(n); } return 0; }