技術標籤：演算法筆記c++學習

[PAT A1069 數字黑洞]

Description:

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 number6174

For example, start from6767, 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 integerNin the range(0,10​4​​).

Output Specification:

If all the 4 digits ofNare the same, print in one line the equationN - N = 0000. Else print each step of calculation in a line until6174comes 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

CODE

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

const int N=4;

int buffer2num(int *buffer,int len){
    int num=0;
    for(int i=0;i<len;i++){
        num=num*10+buffer[i];
    }
    return num;
}

void num2buffer(int *buffer,int len,int num){
    for(int i=len-1;i>=0;i--){
        buffer[i]=num%10;
        num/=10;
    }
}

void print_buffer(int *buffer,int len){
    for(int i=0;i<len;i++)
        printf("%d",buffer[i]);
}

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

int main(){
    int buffer[N]={0};
    int num=0;
    scanf("%d",&num);
    num2buffer(buffer,N,num);
    int res=0;
    while(res!=6174){
        sort(buffer,buffer+N,cmp);
        int a=buffer2num(buffer,N);

        print_buffer(buffer,N);

        printf(" - ");

        sort(buffer,buffer+N);
        int b=buffer2num(buffer,N);
        print_buffer(buffer,N);

        printf(" = ");

        res=a-b;

        num2buffer(buffer,N,res);
        print_buffer(buffer,N);

        printf("\n");
        
        if(res==0)break;
    }
    return 0;
}

注意：

1）類似進位制轉化，轉化程式碼要熟悉

2）傻逼了，列印的時候可以用格式化列印：printf("%04d - %04d = %04d\n",a,b,res);