hdu 1502 Regular Word

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1502

题目描述：

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

23

Sample Output

542

题目大意：

第一行给你一个整数n，代表abc三个字母每个有n个，现在问你有多少种排列方法，使得前任意项都有a的数目>=b的数目>=c的数目

题目分析;

状态转移方程：dp[x][y][z]=dp[x-1][y][z]+dp[x][y-1][z]+dp[x][y][z-1]

Ac代码：

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>using namespace std;char num[65][105],dp[65][65][65][105];void sum(char *a,char *b){    int l1=a[0],l2=b[0],lmax;//a[0]，b[0]代表字符串长度    a[0]=lmax=max(l1,l2);    for(int i=1;i<=lmax;i++)//高精度    {        a[i]+=b[i];        if(a[i]>=10)        {            a[i+1]++;            a[i]%=10;            if(i+1>lmax)            {                a[0]++;                lmax++;            }        }    }}void cpy(char *a,char *b){    for(int i=0;i<=b[0];i++)        a[i]=b[i];}void LCS(int x,int y,int z){       //状态转移方程    if(x>=1&&x-1>=y&&y>=z)        sum(dp[x][y][z],dp[x-1][y][z]);    if(y>=1&&x>=y-1&&y-1>=z)        sum(dp[x][y][z],dp[x][y-1][z]);    if(z>=1&&x>=y&&y>=z-1)        sum(dp[x][y][z],dp[x][y][z-1]);}void Init(){    memset(dp,0,sizeof(dp));    dp[0][0][0][0]=1;    dp[0][0][0][1]=1;    for(int i=1;i<=60;i++)    {        for(int j=0;j<=i;j++)        {            for(int k=0;k<=j;k++)            {                LCS(i,j,k);                if(i==j&&j==k)                    cpy(num[i],dp[i][j][k]);            }        }    }}int main(){    Init();    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=num[n][0];i>=1;i--)            printf("%d",num[n][i]);        printf("\n\n");    }    return 0;}