【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 1:6767Sample Output 1:
7766 - 6677 = 1089 9810 - 0189 = 9621 9621 - 1269 = 8352 8532 - 2358 = 6174Sample Input 2:
2222Sample Output 2:
2222 - 2222 = 0000
分析：
题目简单，但是有很多坑；比如：
（1）
输入2222
2222 - 2222 = 0000
但是输入 222
（2）输入 10000
所以在循环过程中需要不断判断，所得到的的数是否由一个数字构成。
代码：
#include<iostream>
#include<string>
#include<string.h>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
int main(){
char input[7];
cin>>input;
int i,j;
vector<int> vec;
set<int> s;
for(i=0; i<strlen(input); i++){
vec.push_back( input[i] - '0' );
s.insert(input[i] - '0');//利用set自动去重
}
while(vec.size()<4){
vec.push_back(0);
s.insert(0);
}
//排序
sort(vec.begin(),vec.end());
int a=0,b=0;
int t = 10;
for(i=0,j=vec.size()-1; i<vec.size() && j>=0; i++,j--){
a = a*10 + vec[i];
b = b*10 + vec[j];
}
//如果输入的数字仅由一个数字构成
if(s.size() == 1 ){
printf("%04d - %04d = 0000\n",a,a);
return 0;
}
while( (b-a) != 6174){
int t = b-a;
printf("%04d - %04d = %04d\n",b,a,t);
vector<int> v;
while(t != 0){
v.push_back( t%10 );
t /= 10;
}
while(v.size() < 4){
v.push_back(0);
}
sort(v.begin(),v.end());
a = 0;
b = 0;
set<int> set_temp;
for(i=0,j=v.size()-1; i<v.size() && j>=0; i++,j--){
a = a*10 + v[i];
b = b*10 + v[j];
set_temp.insert(v[i]);
}
if(set_temp.size() == 1){
printf("%04d - %04d = 0000\n",a,a);
return 0;
}
}
printf("%04d - %04d = %04d\n",b,a,b-a);
return 0;
}
解法二：
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <set>
using namespace std;
bool cmp(int a, int b){
return a>b;
}
int vec2int(vector<int> vec){
int i,total=0;
for(i=0; i<vec.size(); i++){
total = total*10 + vec[i];
}
return total;
}
void int2vec(int s, vector<int> &vec){
int i;
while(s!=0){
vec.push_back(s%10);
s/=10;
}
while(vec.size()<4){
vec.push_back(0);
}
sort(vec.begin(), vec.end(), cmp);
}
int main(int argc, char** argv) {
string str;
cin>>str;
vector<int> vec;
int i;
for(i=0; i<str.size(); i++){
vec.push_back(str[i]-'0');
}
if(vec.size()==4 && vec[0]==vec[1] && vec[0]==vec[2] && vec[0]==vec[3]){
cout<<str<<" - ";
reverse(str.begin(), str.end());
cout<<str<<" = "<<"0000"<<endl;
return 0;
}
while(vec.size()<4){
vec.push_back(0);
}
sort(vec.begin(), vec.end(), cmp);
int a, b, result;
while(result != 6174){
a = vec2int(vec);
reverse(vec.begin(), vec.end());
b = vec2int(vec);
result = a-b;
printf("%04d - %04d = %04d\n",a,b,result);
vec.clear();
int2vec(result,vec);
}
return 0;
}