Sequence with Digits

最新推荐文章于 2021-09-04 12:35:54 发布

原创最新推荐文章于 2021-09-04 12:35:54 发布 · 804 阅读

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#c++ #c语言 #c#

本文介绍了一种递归数列的计算方法，该方法通过定义数列的每一项为前一项加上最小和最大数位的乘积来生成数列。文章详细解释了如何通过给定的初始值和目标项数来计算数列的特定项。

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);
 }
}