hdu 1502 Regular Words

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

思路：给定一个n，分别由n个a,b,c组成的字符串的所有前缀中a的个数大于等于b的个数大于等于c的个数，求满足这一条件的字符串的个数有多少，本题要用到dp及大数运算，转移方程：dp[i][j][k]=dp[i-1][j][k]+dp[i][j-1][k]+dp[i][j][k-1],其中i,j,k分别代表a,b,c的个数

老实说并不会写大数，不过发现了一段很神奇的代码

 #include<iostream>
 #include<iomanip>
 #include<cstdio>
 #include<cstring>
 #include<sstream>
 #include<stack>
 #include<queue>
 #include<fstream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<vector>
 #include<cmath>
 #define CLR(x) memset(x,0,sizeof(x))
 #define ll long long
 #define eps 1e-6
 #define pow2(x) ((x)*(x))
 #define forto(i,n) for(int i=0;i<n;i++)
 #define for1to(i,n) for(int i=1;i<=n;i++)
 #define VI vector<int>
 using namespace std;
 const double PI=acos(-1.0);
 #define INF 0x3fffffff          ///INT_MAX一半防止上溢
 #define NINF 0xbfffffff
 
 int dp[][][][];
 
 int add(int A[],int B[])        ///Add B to A
 {
     for (int i=;i<;i++)
     {
         A[i]+=B[i];
         A[i+]+=A[i]/;
         A[i]%=;
     }
 }
 
 int main()
 {
     dp[][][][]=;
     for (int i=;i<=;i++)
         for (int j=;j<=i;j++)
             for (int k=;k<=j;k++)
             {
                 add(dp[i][j][k],dp[i-][j][k]);
                 add(dp[i][j][k],dp[i][j-][k]);
                 add(dp[i][j][k],dp[i][j][k-]);
             }
 
     int n;
     while (cin>>n)
     {
         int k();
         while (!dp[n][n][n][k])
             k--;
         for (int i=k;i>=;i--)
         {
             if (i!=k)
                 cout<<setw()<<setfill('');
             cout<<dp[n][n][n][i];
         }
         cout<<"\n"<<endl;
     }
     return ;
 } 

还有打表的代码

 #include<iostream>
 #include<stdio.h>
 #include<string.h>
 using namespace std;
 char s[][]={
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 "",
 };
 int main()
 {
     int n;
     while(scanf("%d",&n)>)
     {
         printf("%s\n\n",s[n]);
     }
     return ;
 }