UNIMODAL PALINDROMIC DECOMPOSITIONS OpenJ_Bailian - 1221

转载于 2021-08-02 17:36:24 发布 · 871 阅读

CC 4.0 BY-SA版权

原文链接：https://blog.youkuaiyun.com/qq_37451344/article/details/81094022?utm_medium=distribute.pc_relevant.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu%7Edefault-10.control&depth_1-utm_source=distribute.pc_relevant.none-task-blog-2%7Edefault%7EBlogCommendFromBaidu

算法刷题同时被 2 个专栏收录

203 篇文章

订阅专栏

codeforce

168 篇文章

订阅专栏

本文介绍了一种特定类型的回文序列——单峰回文序列，并探讨了如何计算一个整数的所有可能单峰回文分解的数量。通过动态规划的方法，提供了一个有效的解决方案。

A sequence of positive integers is Palindromic if it reads the same forward and backward. For example:
23 11 15 1 37 37 1 15 11 23
1 1 2 3 4 7 7 10 7 7 4 3 2 1 1
A Palindromic sequence is Unimodal Palindromic if the values do not decrease up to the middle value and then (since the sequence is palindromic) do not increase from the middle to the end For example, the first example sequence above is NOT Unimodal Palindromic while the second example is.
A Unimodal Palindromic sequence is a Unimodal Palindromic Decomposition of an integer N, if the sum of the integers in the sequence is N. For example, all of the Unimodal Palindromic Decompositions of the first few integers are given below:
1: (1)
2: (2), (1 1)
3: (3), (1 1 1)
4: (4), (1 2 1), (2 2), (1 1 1 1)
5: (5), (1 3 1), (1 1 1 1 1)
6: (6), (1 4 1), (2 2 2), (1 1 2 1 1), (3 3),
(1 2 2 1), ( 1 1 1 1 1 1)
7: (7), (1 5 1), (2 3 2), (1 1 3 1 1), (1 1 1 1 1 1 1)
8: (8), (1 6 1), (2 4 2), (1 1 4 1 1), (1 2 2 2 1),
(1 1 1 2 1 1 1), ( 4 4), (1 3 3 1), (2 2 2 2),
(1 1 2 2 1 1), (1 1 1 1 1 1 1 1)

Write a program, which computes the number of Unimodal Palindromic Decompositions of an integer.

Input

Input consists of a sequence of positive integers, one per line ending with a 0 (zero) indicating the end.

Output

For each input value except the last, the output is a line containing the input value followed by a space, then the number of Unimodal Palindromic Decompositions of the input value. See the example on the next page.

Sample Input

Sample Output

2 2
3 2
4 4
5 3
6 7
7 5
8 11
10 17
23 104
24 199
131 5010688
213 1055852590
92 331143

Hint

N < 250


#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll long long
#define maxn 305
#define ms memset
using namespace std;
/*
题目大意：定义一个新的回文数字串，
要求其和要一样都是n，并且从两边到中间的
数字段是不减的，对于给定的n，求计数。
定义dp[i][j],即和为i的两端为j组成的回文串计数。
枚举两端的j,对于剩余的i-2*j,累加其dp[i-2*j][k],
其中k是从j到i-2*j的范围。
最后dp[i][i]特殊化为1.
*/
ll n,table[maxn][maxn];
void init()
{
    memset(table,0,sizeof(table));
    for(int i=0;i<maxn;i++) table[0][i]=1;///多余
    for(int i=1;i<maxn;i++)
    {
         table[i][1]=0;
         for(int j=1;j<=i/2;j++)
         {
             if(i-2*j==0)
             {
                 table[i][j]=1;
                 break;
             }
             for(int k=j;k<=i-2*j;k++)
               table[i][j]+=table[i-2*j][k];
         }
        table[i][i]=1;
    }
}
 
int main()
{
    init();
    while( scanf("%lld",&n) &&n )
    {
        ll ans=0;
        for(int i=0;i<=n;i++) ans+=table[n][i];
        printf("%lld %lld\n",n,ans);
    }
    return 0;

}