HDU_1502_dp

Regular Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2102    Accepted Submission(s): 825

Problem Description

Consider words of length 3n over alphabet {A, B, C} . Denote the number of occurences of A in a word a as A(a) , analogously let the number of occurences of B be denoted as B(a), and the number of occurenced of C as C(a) .

Let us call the word w regular if the following conditions are satisfied:

A(w)=B(w)=C(w) ; 
if c is a prefix of w , then A(c)>= B(c) >= C(c) . 
For example, if n = 2 there are 5 regular words: AABBCC , AABCBC , ABABCC , ABACBC and ABCABC .

Regular words in some sense generalize regular brackets sequences (if we consider two-letter alphabet and put similar conditions on regular words, they represent regular brackets sequences).

Given n , find the number of regular words.

 

Input

There are mutiple cases in the input file.

Each case contains n (0 <= n <= 60 ).

There is an empty line after each case.

 

Output

Output the number of regular words of length 3n .

There should be am empty line after each case.

 

Sample Input

2

3

 

Sample Output

5

42

 

还是看题解做的，蓝瘦。。。

dp[i][j][k][100]; i这维表示a的个数，j这维表示b的个数，k这维表示c的个数。

dp过程中需要用大数加法。

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

void add(char A[],char B[],char C[])
{
    int a=;
    int len1=strlen(A);
    int len2=strlen(B);
    int wei=;
    while(a<len1&&a<len2)
    {
        int sum=A[a]+B[a]-''-''+wei;
        wei=;
        if(sum>)
        {
            wei++;
            sum-=;
        }
        C[a]=sum+'';
        a++;
    }
    while(a<len1)
    {
        int sum=A[a]+wei-'';
        wei=;
        if(sum>)
        {
            wei++;
            sum-=;
        }
        C[a]=sum+'';
        a++;
    }
    while(a<len2)
    {
        int sum=B[a]+wei-'';
        wei=;
        if(sum>)
        {
            wei++;
            sum-=;
        }
        C[a]=sum+'';
        a++;
    }
    if(wei>)
        C[a++]='';
    C[a]='\0';
}

char dp[][][][];

int main()
{
    char A[],B[],C[];
    for(int i=; i<=; i++)
        for(int j=; j<=; j++)
            for(int k=; k<=; k++)
                strcpy(dp[i][j][k],"");
    strcpy(dp[][][],"");

    for(int i=; i<=; i++)
        for(int j=; j<=; j++)
            for(int k=; k<=; k++)
            {
                if(i>=j&&j>=k)
                {
                    add(dp[i-][j][k],dp[i][j-][k],dp[i][j][k]);
                    add(dp[i][j][k],dp[i][j][k-],dp[i][j][k]);
                }
            }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int len=strlen(dp[n][n][n]);
        char res[];
        for(int i=len-;i>=;i--)
            res[len-i-]=dp[n][n][n][i];
        res[len]='\0';
        printf("%s\n\n",res);
        //getchar();
    }
    return ;
}