hdu 5062 Beautiful Palindrome Number( dp )_palipali邀请码-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44259177

本文提供了一种方法来计算指定范围内中心向两侧递减的回文数的数量，通过动态规划实现效率优化。

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Beautiful Palindrome Number

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 474 Accepted Submission(s): 301

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

Sample Output

  
   9
258

Source

BestCoder Round #13
题目大意:求取范围内的中心向两侧递减的回文数的数量
题目分析:dp求取递减串的个数
dp[i][j]:表示长度为i的以j为最大项的递减/递增串的个数
然后求和即可,
特判一下,输入0的时候输出1即可
数据水到,好像各种姿势都能过

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

int t,n;
int num[10];
int dp[10][10];

int main ( )
{
    scanf ( "%d" , &t );
    memset ( dp , 0 , sizeof ( dp ) );
    for ( int i = 1; i < 10 ; i++ )
        dp[1][i] = 1;
    for ( int i = 2 ; i < 10 ; i++ )
        for ( int j = 1 ; j < 10 ;j++ )
            for ( int k = 1 ; k < j ; k++ )
                dp[i][j] += dp[i-1][k];
    num[0] = 0;
    for ( int i = 1 ; i < 7 ; i++ )
    {
        int mid = (i+1)>>1;
        num[i] += num[i-1];
        for ( int j = 1; j < 10 ; j++ )
            num[i] += dp[mid][j];
    }
    while ( t-- )
    {
        scanf ( "%d" , &n );
        if ( n == 0 ) puts ( "1" );
        else printf ( "%d\n" , num[n] );
    }
}