Sequence with Digits

2023-03-14 16:14 8阅读 0赞

Let’s define the following recurrence : an+1=an+minDigit(an)⋅maxDigit(an).an+1=an+minDigit(an)⋅maxDigit(an).Here minDigit(x)minDigit(x) and maxDigit(x)maxDigit(x) are the minimal and maximal digits in the decimal representation of x without leading zeroes.  
For examples refer to notes.Your task is calculate aK for given a1 and K.  
Input  
The first line contains one integer t(1≤t≤1000) — the number of independent test cases.  
Each test case consists of a single line containing two integers a1 and K (1≤a1≤1018, 1≤K≤1016) separated by a space.  
Output  
For each test case print one integer aKaK on a separate line.  
Example  
input  
8  
1 4  
487 1  
487 2  
487 3  
487 4  
487 5  
487 6  
487 7  
ootput  
42  
487  
519  
528  
544  
564  
588  
628  
Note  
a1=487  
a2=a1+minDigit(a1)⋅maxDigit(a1)=487+min(4,8,7)⋅max(4,8,7)=487+4⋅8=519  
a3=a2+minDigit(a2)⋅maxDigit(a2)=519+min(5,1,9)⋅max(5,1,9)=519+1⋅9=528  
a4=a3+minDigit(a3)⋅maxDigit(a3)=528+min(5,2,8)⋅max(5,2,8)=528+2⋅8=544  
a5=a4+minDigit(a4)⋅maxDigit(a4)=544+min(5,4,4)⋅max(5,4,4)=544+4⋅5=564  
a6=a5+minDigit(a5)⋅maxDigit(a5)=564+min(5,6,4)⋅max(5,6,4)=564+4⋅6=588  
a7=a6+minDigit(a6)⋅maxDigit(a6)=588+min(5,8,8)⋅max(5,8,8)=588+5⋅8=628

#include<stdio.h>
    int fmax(int n)
    { 
     int res=n%10;
     while(n)
     { 
      if(res<n%10)
      { 
       res=n%10;
      }
      n=n/10;
     }
     return res;
    }
    int fmin(int n)
    { 
     int res=n%10;
     while(n)
     { 
      if(res>n%10)
      { 
       res=n%10;
      }
      n=n/10;
     }
     return res;
    }
    int fmn(int n)
    { 
     return n+fmax(n)*fmin(n);
    }
    int main()
    { 
     int t=0;
     scanf("%d",&t);
     while(t--)
     { 
      int a1,k=0;
      scanf("%d%d",&a1,&k);
      for(int i=1;i<k;i++)
      { 
       a1=fmn(a1);
      }
      printf("%d\n",a1);
     }
    }