PAT 甲级 1069 The Black Hole of Numbers (20 分)(内含别人string处理的精简代码)

1069 The Black Hole of Numbers (20 分)  

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 (.

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

作者: CHEN, Yue 单位: 浙江大学 时间限制: 200 ms 内存限制: 64 MB 代码长度限制: 16 KB

 

题意:

给一个数组n，先对各个数字位从大到小排序得到a，然后逆置得到b，计算a-b，得到c，重复上述过程，直到得到6174，或者，对于特例，四个数字位完全相同，那么输出0000。

题解:

刚开始输入的数字可能不足四位要自动补0 以后的结果可能也不足四位都要自动补0

结果是0或者6174都是要退出的

一开始没考虑到6174，测试点5没过，后来想到了

AC代码:

#include<iostream>
#include<algorithm>
using namespace std;
int n;
int a[10];
int main(){
     cin>>n;
     int big=0;
     int small=0;
     int x=n;
     int k=0,f;
     if(x==6174){
        printf("7641 - 1467 = 6174");
        return 0;
     }
     while(x!=6174){
         k=0;
         big=0;
         small=0;
         while(x>=10){
             a[++k]=x%10;
             x/=10;
         }
         a[++k]=x;
         while(k<4){
             a[++k]=0;
         }
         sort(a+1,a+1+4);
         for(int i=1;i<=4;i++){
             big=big*10+a[5-i];
             small=small*10+a[i];
         }
         x=big-small;
         printf("%04d - %04d = %04d\n",big,small,x);
         if(x==0) break;
     }
     return 0;
}
     

 

更简洁的string处理的代码:

#include<bits/stdc++.h>
using namespace std;
bool cmp(char a,char b) 
{
    return a>b;
}
int main(void)
{
    string s;
    cin>>s;
    s.insert(0,4-s.size(),'0');
    do{
        string a=s,b=s;
        sort(a.begin(),a.end(),cmp);
        sort(b.begin(),b.end());
        int diff=stoi(a)-stoi(b);
        s=to_string(diff);
        s.insert(0,4-s.size(),'0');
        cout<<a<<" - "<<b<<" = "<<s<<endl;
    }while(s!="6174"&&s!="0000");
    return 0;

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版权声明：本文为CSDN博主「Imagirl1」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/Imagirl1/article/details/82261213

 

标签：1069,digit,PAT,string,int,number,6174,9621
来源： https://www.cnblogs.com/caiyishuai/p/11770567.html