PAT 甲級 1069 The Black Hole of Numbers (20 分)(內含別人string處理的精簡代碼)

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

做者: CHEN, Yue

單位: 浙江大學

時間限制: 200 ms

內存限制: 64 MB

代碼長度限制: 16 KB

 

題意:

給一個數組n，先對各個數字位從大到小排序獲得a，而後逆置獲得b，計算a-b，獲得c，重複上述過程，直到獲得6174，或者，對於特例，四個數字位徹底相同，那麼輸出0000。

題解:

剛開始輸入的數字可能不足四位要自動補0 之後的結果可能也不足四位都要自動補0

結果是0或者6174都是要退出的

一開始沒考慮到6174，測試點5沒過，後來想到了

AC代碼:

#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;
}
     

 

更簡潔的string處理的代碼:

#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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版權聲明：本文爲CSDN博主「Imagirl1」的原創文章，遵循 CC 4.0 BY-SA 版權協議，轉載請附上原文出處連接及本聲明。
原文連接：https://blog.csdn.net/Imagirl1/article/details/82261213