hdu 5062 Beautiful Palindrome Number（模拟）

Beautiful Palindrome Number

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1062    Accepted Submission(s): 688

Problem Description

A positive integer x can represent as  (a1a2…akak…a2a1)10  or  (a1a2…ak−1akak−1…a2a1)10  of a 10-based notational system, we always call x is a Palindrome Number. If it satisfies  0<a1<a2<…<ak≤9 , we call x is a Beautiful Palindrome Number.
Now, we want to know how many Beautiful Palindrome Numbers are between 1 and  10N .

 

Input

The first line in the input file is an integer  T(1≤T≤7) , indicating the number of test cases.
Then T lines follow, each line represent an integer  N(0≤N≤6) .

 

Output

For each test case, output the number of Beautiful Palindrome Number.

 

Sample Input

  
  

   
   2
1
6
  
  

 

Sample Output

  
  

   
   9
258
  
  
  
  


  
  

题意：1~10的n次方有几个上升的回文数

思路：直接模拟，当然像我这样打表也行，n只有6

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
/*int a[10];
int check(int x)
{
    int l=x,ans=0;
    int num=0;
    while(l)
    {
        ans=ans*10+l%10;
        l/=10;
        num++;
    }
    if(ans!=x) return 0;
    if(num&1) num++;
    num/=2;
    int now=-1;
    while(num--)
    {
        int w=x%10;
        if(w<=now) return 0;
        now=w;
        x/=10;
    }
    return 1;
}
void init()
{
    int num=0;
    for(int i=1;i<=100;i++)
    {
        if(check(i))
            num++;
    }
    printf("%d\n",num);
}*/
int main()
{
    int T,n;
    //init();
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        if(n==0) printf("1\n");
        else if(n==1) printf("9\n");
        else if(n==2) printf("18\n");
        else if(n==3) printf("54\n");
        else if(n==4) printf("90\n");
        else if(n==5) printf("174\n");
        else printf("258\n");
    }
    return 0;
}