pat1069（误解的代价）

1069. 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 = 10899810 - 0189 = 96219621 - 1269 = 83528532 - 2358 = 6174

Sample Input 2:

2222

Sample Output 2:

2222 - 2222 = 0000

一开始就把题意理解错了，应该是输入整型数字，偏偏自以为是认识输入的是字符串，结果错了错了错了！！！转换了好久的说，最后也没全通过，引以为戒，不过学到了字符和数字转换

#include<iostream> #include<string>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;bool cmp(char a,char b){if(a>b){return true;}else{return false;}}int change1(string a){int ans=0;sort(a.begin(),a.end());for(int i=a.size()-1;i>=0;i--){ans+=(a[i]-'0')*(pow(10,(a.size()-1-i)));}return ans;}int change2(string a){int ans=0;sort(a.begin(),a.end(),cmp);for(int i=a.size()-1;i>=0;i--){ans+=(a[i]-'0')*(pow(10,(a.size()-1-i)));}return ans;}int main(){int tt;cin>>tt; string s1;char ss[10]={};//itoa(tt,ss,10);sprintf(ss,"%d",tt);s1=ss;int sub=0;int t1=0,t2=0;while(sub!=6174){t1=change1(s1);t2=change2(s1);sub=t2-t1;if(sub==0){//cout<<t2<<" - "<<t1<<" = "<<sub<<endl;printf("%04d - %04d = %04d\n",t2,t1,sub);return 0;}else{//cout<<t2<<" - "<<t1<<" = "<<sub<<endl;printf("%04d - %04d = %04d\n",t2,t1,sub);char s2[10]={};//itoa(sub,s2,10);sprintf(s2,"%d",sub);//数字转成字符串 s1=s2;}}//cout<<t2<<" - "<<t1<<" = "<<sub<<endl;return 0;}