V - Regular Words
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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.
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.
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.
There should be am empty line after each case.
Sample Input
2
3
Sample Output
5
42
一开始我以为只是简单的dp,感觉怎么这么简单啊,就随便写了一下提交了,然后就wa了,又检查了一下,觉得没什么问题,无奈只好求助于网上的题解,才知道原来是高精度加法,我就哭了! 我只写过杭电的1002,那和这里的高精度是一个档次的吗?估计了一下,这道题写了整整一天,各种改错,实在是没办法,问题太多了。。。。。。看来是字符串处理水平不足啊...话不多说,从dp的角度看,这道题其实不难,注意别超时,一开始用了四层for循环,180*60*60*60觉得不会超时,TLE了以后反应过来原来加法过程还要乘以一个数,这样就超时了,优化一下变成三重循环,这才勉强符合时限;因为没写过类似的高精度加法,所以写起来特别费力,字符串结束符号那里我估计是最让我蛋疼的地方,然后是一开始不知道怎么给字符串赋值,再就是我居然连高精度数都没搞清楚,还好学长提醒了我一下,这才少走了许多歪路,在这里感谢cxlove对我的提点,不然我估计就要写两天了。。。囧。。。
代码如下:
(代码写的有点搓,看来以后还得学点艺术啊。。。)
#include<stdio.h> #include<string.h> char dp[61][61][61][100]; void sum(char a[100],char b[100]) { long z=0,i=0,j=0; char c; while(a[i]!=''&&b[j]!='') { c=b[j]; b[j]=(a[i]+b[j]-'0'-'0'+z)%10+48; z=(a[i]+c-'0'-'0'+z)/10; i++; j++; } if(a[i]!=''&&b[j]=='') { while(a[i]!='') { b[j]=(a[i]+z-'0')%10+48; z=(a[i]+z-'0')/10; i++; j++; } if(z!=0) { b[j]=z+'0'; b[j+1]=''; } else b[j]=''; } else { while(a[i]==''&&b[j]!='') { c=b[j]; b[j]=(b[j]+z-'0')%10+48; z=(c+z-'0')/10; j++; } if(z!=0) { b[j]=z+'0'; b[j+1]=''; } else b[j]=''; } } int max(long a,long b) { return a>b?a:b; } int min(long a,long b) { return a>b?b:a; } int main() { long n,i,j,k,l,m; while(scanf("%ld",&n)!=EOF) { for(j=0; j<=n; j++) for(k=0; k<=n; k++) for(l=0; l<=n; l++) { dp[j][k][l][0]='0'; dp[j][k][l][1]=''; } for(j=1; j<=n; j++) for(k=0; k<=j; k++) for(l=0; l<=k; l++) { if(k==0) { dp[j][k][l][0]='1'; dp[j][k][l][1]=''; } else if(l==0) { sum(dp[j-1][k][l],dp[j][k][l]); sum(dp[j][k-1][l],dp[j][k][l]); } else { sum(dp[j-1][k][l],dp[j][k][l]); sum(dp[j][k-1][l],dp[j][k][l]); sum(dp[j][k][l-1],dp[j][k][l]); } } l=strlen(dp[n][n][n]); for(j=l-1; j>=0; j--) printf("%c",dp[n][n][n][j]); printf(" "); printf(" "); } return 0; }