HDU1502 动态规划计数 高精度

Regular Words

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 379    Accepted Submission(s): 169

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 thenumber of occurences of B be denoted as B(a), and the number of occurenced of Cas 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 weconsider two-letter alphabet and put similar conditions on regular words, theyrepresent 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

 

动态规划计数+高精度,不懂写Java大数,杯具了好久.

 

#include<stdio.h>#include<stdlib.h>#include<string.h>#define N 62#define LL long longchar dp[N][N][N][100],res[100];int min(int a,int b){ return a<b?a:b;}int max(int a,int b){ return a>b?a:b;}void add(char a[],char b[]){ int lena,lenb,t,i,j,l; lena=strlen(a); lenb=strlen(b); char c[100]; l=max(lena,lenb)+2; memset(c,0,sizeof(c)); t=-1; i=lena-1; j=lenb-1; while(i>=0 && j>=0) { c[++t]=a[i]+b[j]-'0'; i--; j--; } if (i>=0) while (i>=0) { c[++t]=a[i]; i--; }elsewhile (j>=0){c[++t]=b[j];j--;}for (i=0; i<=t; ++i){if (c[i]-'0'>=10){c[i]-=10;if (i!=t) c[i+1]++;else{c[i+1]='1';++t;}}}memset(a,0,sizeof(a));for (i=0; i<=t; ++i) a[i]=c[t-i];a[t+1]='/0';}int main(){ int n,len,i,j,k,l,a,b,tt;strcpy(dp[0][0][0],"1/0");strcpy(dp[1][0][0],"1/0");strcpy(dp[0][1][0],"0/0"); strcpy(dp[0][0][1],"0/0"); strcpy(dp[1][1][0],"1/0");strcpy(dp[0][1][1],"0/0"); strcpy(dp[1][0][1],"0/0");for (i=1; i<=60; ++i){for (j=0; j<=i; ++j){for (k=0;k<=j;++k){strcpy(res,"0/0"); if (i>=j && j>=k) { if (i>j && j>=k) add(res,dp[i-1][j][k]); if (j>k && i>=j) add(res,dp[i][j-1][k]); if (k>0 && i>=j && j>=k) add(res,dp[i][j][k-1]); }strcpy(dp[i][j][k],res);}}} while (scanf("%d",&n)!=EOF) { if (n==0) { printf("0/n/n"); continue; } puts(dp[n][n][n]); printf("/n"); } return 0;}