主题链接：http://acm.hdu.edu.cn/showproblem.php?
pid=5062
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.
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).
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
Source
题意：
求1到10的n次方的范围内，满足：
1、是回文数；
2、回文的前半部分满足升序。
打表代码：
#include <cstdio>
#include <cstring>
int find_num(int num)
{
int a[17];
memset(a,0,sizeof(a));
int L = 0;
while(num)
{
a[++L] = num%10;
num/=10;
}
for(int i = 1; i <= L/2; i++)//回文
{
if(a[i] != a[L-i+1])
return 0;
}
for(int i = 1; i < L/2+L%2; i++)//升序
{
if(a[i+1] <= a[i])
return 0;
}
return 1;
}
int main()
{
int sum[17];
memset(sum,0,sizeof(sum));
sum[0] = 1;
for(int i = 1; i <= 1000000; i++)
{
int flag = find_num(i);
if(flag)
{
//printf("num::%d\n",i);
if(i <= 10)
sum[1]++;
if(i <= 100)
sum[2]++;
if(i <= 1000)
sum[3]++;
if(i <= 10000)
sum[4]++;
if(i <= 100000)
sum[5]++;
if(i <= 1000000)
sum[6]++;
}
}
int t;
for(int i = 0; i <= 6; i++)
{
printf("%d::%d\n",i,sum[i]);
}
return 0;
}
代码例如以下：
#include <cstdio>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
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 if(n == 6)
printf("258\n");
}
return 0;
}
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