hdu 5062 Beautiful Palindrome Number（Bestcodeer Round #13）

Beautiful Palindrome Number

                                                                Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                Total Submission(s): 116    Accepted Submission(s): 82

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

216

 

Sample Output

9258

 

Source

BestCoder Round #13

 

结果就只有七个，交上去一个表就行了，n最大为6，暴力代码也能过。附上打表代码和暴力代码。

打表代码：

//0ms#include <cstdio>#include <cstring>#include <iostream>int a[7]={1,9,18,54,90,174,258};int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        printf("%d\n",a[n]);    }}

暴力代码：

//46ms#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[10];bool judge(int x){    int n=0;    while(x>0)    {        a[++n]=x%10;        x=x/10;    }    //printf("%d\n",n);    if(a[1]!=a[n])    return false;    for(int j=2,k=n-1;j<=k;j++,k--)    {        if(a[j]>a[j-1]&&a[j]==a[k])        ;        else        return false;    }    return true;}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int temp=1;        int ans=0;        for(int i=1;i<=n;i++)        {            temp*=10;        }        //printf("%d",temp);        for(int i=1;i<=temp;i++)        {            if(judge(i))            {                ans++;            }        }        printf("%d\n",ans);    }    return 0;}