1069. The Black Hole of Numbers (20)

1. 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

---

只要建立数字和数组的关系即可。注意开始要输入到数字里，不然若输入不满4位数，输入到数组可能会出错。

---

#include<iostream>
#include<algorithm>
using namespace std;
int getnum(char num[]){
  int t=0;
  for(int i=0;i<=3;i++){
      t=t*10+(num[i]-'0');
  }
  return t;
}
void numtstr(int n,char *str){
    for(int i=3;i>=0;i--){
        str[i]=n%10+'0';
        n/=10;
    }
}
int reverse(int n){
    int t=0;
    for(int i=0;i<4;i++){
        t=t*10+n%10;
        n/=10;
    }
    return t;
}
bool cmp(char a,char b){
    return a>b;
}
int main(){
    int n,n1,t;
    char num[4]={0};
    scanf("%d",&t);
    numtstr(t,num);
    sort(num,num+4,cmp);
    n=getnum(num);
    while(1){
        n1=reverse(n);
        t=n-n1;
        printf("%04d - %04d = %04d\n",n,n1,t);
        if(t==0||t==6174) break;
        numtstr(t,num);
        sort(num,num+4,cmp);
        n=getnum(num);
    }
    return 0;
}

（以前写的）用节点写的，实际上不需要建立结点，直接基于数组进行运算就好。

#include<iostream>
#include<algorithm>
using namespace std;

bool compare(int a,int b)
{
    return a>b;
}

struct node
{
public:
    int a[4];
    node(){}
    int m;
    node(int n)
    {
        a[0]=n/1000;
        a[1]=(n/100)%10;
        a[2]=(n/10)%10;
        a[3]=(n%10);
        sort(a,a+4,compare);
    }
    node(int b[])
    {
        a[0]=b[3];
        a[1]=b[2];
        a[2]=b[1];
        a[3]=b[0];
    }
    node(int de[],int in[])
    {

        /*
        a[0]=de[0]-in[0];
        a[1]=de[1]-in[1];
        a[2]=de[2]-in[2];
        a[3]=de[3]-in[3];
        */
        int s1=de[0]*1000+de[1]*100+de[2]*10+de[3];
        int s2=in[0]*1000+in[1]*100+in[2]*10+in[3];
        int n=s1-s2;
        a[0]=n/1000;
        a[1]=(n/100)%10;
        a[2]=(n/10)%10;
        a[3]=(n%10);
        m=n;
        sort(a,a+4,compare);
    }
    void pirntm()
    {
        printf("%d%d%d%d",  m/1000,(m/100)%10
        ,(m/10)%10
        ,(m%10));
    }
    void printnum()
    {
        printf("%d%d%d%d",a[0],a[1],a[2],a[3]);
    }
    int check()
    {
        int s=a[0]*1000+a[1]*100+a[2]*10+a[3];
        if(m==0)    
            return 0;
        else if(m==6174)
            return 1;
        else return -1;
    }
};
int main()
{
    int n;
    cin>>n;
    //n=1231;
    node s1=node(n);
    node s2=node(s1.a);
    while(s1.check()==-1)
    {
        node s=node(s1.a,s2.a);
        s1.printnum();cout<<" - ";s2.printnum();cout<<" = ";s.pirntm();cout<<endl;
        s1=s;
        s2=node(s1.a);
    }
    system("pause");
    return 0;
}