HDU 5062 Beautiful Palindrome Number(数学题)

题目链接：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.

 

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

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;}

代码如下：

#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;}