hdu5062——Beautiful Palindrome Number（模拟）

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

输入只有其中情况，可以事先算出来，直接把答案存到数组里

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <math.h>
#include <algorithm>
#include <queue>
#include <iomanip>
#define INF 0x3f3f3f3f
#define MAXN 6005
#define Mod 99999999
using namespace std;
int main()
{
/*for(int n=0; n<=6; ++n)
{
int len=pow(10,n),ans=0;
for(int i=1; i<=len; ++i)
{
if(i>=1&&i<=9)
ans++;
else if(i>=10&&i<=99)
{
int a=i/10,b=i%10;
if(a==b)
{
ans++;
//cout<<i<<endl;
}
}
else if(i>=100&&i<=999)
{
int a=i/100,b=(i/10)%10,c=i%10;
if(a<b&&a==c)
{
ans++;
//cout<<i<<endl;
}
}
else if(i>=1000&&i<=9999)
{
int k1=i/1000,k2=(i/100)%10,k3=(i/10)%10,k4=i%10;
if(k1==k4&&k2==k3&&k1<k2)
{
ans++;
//cout<<i<<endl;
}
}
else if(i>=10000&&i<=99999)
{
int k1=i/10000,k2=(i/1000)%10,k3=(i/100)%10,k4=(i/10)%10,k5=i%10;
if(k1==k5&&k2==k4&&k1<k2&&k2<k3)
{
ans++;
cout<<i<<endl;
}
}
else if(i>=100000&&i<=999999)
{
int k1=i/100000,k2=(i/10000)%10,k3=(i/1000)%10,k4=(i/100)%10,k5=(i/10)%10,k6=i%10;
if(k1==k6&&k2==k5&&k3==k4&&k1<k2&&k3<k4&&k2<k3)
{
ans++;
//cout<<i<<endl;
}
}

}
cout<<n<<" "<<len<<" "<<ans<<endl;
}
return 0;*/
int a[]={1,9,18,54,90,174,258};
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
cout<<a
<<endl;
}
}