.
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.
Each case contains n (0 <= n <= 60 ).
There is an empty line after each case.
There should be am empty line after each case.
2 3
5 42
题意：就是求ABC每个字母有n个，求排列的个数
思路：dp思想，按照题目要求的排列的方式，得出公式dp[i][j][k] = dp[i-1][j][k]+dp[i][j-1][k]+dp[i][j][k-1]，并且i>=j>=k.但是要注意这些数字到了后面将会越来越大，所以必须要用大数模拟
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
char dp[65][65][65][85];
void add(char a[],char b[],char back[])
{
int i,j,k,up,x,y,z,l;
char *c;
if(strlen(a) > strlen(b))
l = strlen(a)+2;
else
l = strlen(b)+2;
c = (char*)malloc(l*sizeof(char));
i = strlen(a)-1;
j = strlen(b)-1;
k = 0;
up = 0;
while(j>=0 || i>=0)
{
if(i<0) x = '0';
else
x = a[i];
if(j<0) y = '0';
else
y = b[j];
z = x-'0'+y-'0';
if(up)
z++;
if(z>9)
{
up = 1;
z%=10;
}
else
up = 0;
c[k++] = z+'0';
i--;
j--;
}
if(up)
c[k++] = '1';
i = 0;
c[k] = '\0';
for(k-=1; k>=0; k--)
back[i++] = c[k];
back[i] = '\0';
}
int main()
{
int n,i,j,k;
for(i = 0; i<=60; i++)
for(j = 0; j<=60; j++)
for(k = 0; k<=60; k++)
strcpy(dp[i][j][k],"0");
strcpy(dp[1][0][0],"1");
strcpy(dp[1][1][0],"1");
strcpy(dp[1][1][1],"1");
for(i = 2; i<=60; i++)
{
for(j = 0; j<=i; j++)
{
for(k = 0; k<=j; k++)
{
if(i-1>=j)
add(dp[i-1][j][k],dp[i][j][k],dp[i][j][k]);
if(j-1>=k)
add(dp[i][j-1][k],dp[i][j][k],dp[i][j][k]);
if(j>=k-1)
add(dp[i][j][k],dp[i][j][k-1],dp[i][j][k]);
}
}
}
while(~scanf("%d",&n))
{
printf("%s\n\n",dp[n][n][n]);
}
return 0;
}