solution Of 1069. The Black Hole of Numbers (20)

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

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结题思路 :
题意要求我们在输出4位数的“黑洞”数字之前，输出求差的过程。
要求1：直接输入“黑洞数”值6174，我们仍然需要进行一步的输出；
要求2：当两个数值的差为6174时停止计算。

程序步骤：
第一步、对于输入数据进行字符化；
第二步、字符串内排序后求差。

具体程序(AC)如下：

#include<iostream>#include<string>#include<cstring>#include<cstdlib>#include<cmath> #include<sstream>#include<algorithm>using namespace std;int cmp(char a,char b){    return a>b;}string opDiff(string a){     int startA,startB;     char tmpA[5];     char tmpB[5];     strcpy(tmpA,a.c_str());     sort(tmpA,tmpA+4,cmp);     for(int i=0;i<4;++i)         tmpB[i]=tmpA[3-i];     tmpB[4]=0;     cout<<tmpA<<" - "<<tmpB<<" = " ;    for(int i=3;i>=0;--i)    {       if(tmpA[i]>=tmpB[i])          tmpA[i]=tmpA[i]-tmpB[i]+'0';       else       {          tmpA[i]=tmpA[i]+10;          tmpA[i-1]=tmpA[i-1]-1;           tmpA[i]=tmpA[i]-tmpB[i]+'0';       }    }    cout<<tmpA<<endl;    a=tmpA;    return a;}bool reversedEqual(char* tmp){    int len=strlen(tmp);    for(int i=0;i<len/2;++i)        if(tmp[i]!=tmp[len-1-i])            return false;    return true;}int main(){    int value;    cin>>value;    char value2Str[5];    sprintf(value2Str,"%04d",value);    if(reversedEqual(value2Str))      cout<<value2Str<<" - "<<value2Str<<" = 0000"<<endl;    else    {        string sF(value2Str);        if(sF=="6174")        {            cout<<"7641 - 1467 = 6174"<<endl;            return 0;        }        while(sF!="6174")            sF=opDiff(sF);    }    return 0;   }