Rabbit Board (hdu2067) and rabbit board hdu2067

Rabbit Board (hdu2067) and rabbit board hdu2067
Rabbit Board

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 7547 Accepted Submission (s): 4020

Problem Description the uncle of the rabbit came back from outside to bring her a gift. The rabbit happily ran back to his room and opened a board. The rabbit was disappointed. But a few days later I found the fun of the Board. The number of shortest paths from the start point (0, 0) to the end point (n, n) is C (2n, n ), now, the rabbit wants to determine the number of such paths without traversing the diagonal line (but can touch the grid points on the diagonal line? I have never thought about it for a long time. Now I want to ask you to help me solve this problem. It should be difficult for you!

 

Input n (1 <= n <= 35), and end the Input when n is equal to-1.

 

For the number of Output paths for each input data, see Sample for the specific format.

 

Sample Input1312-1

 

Sample Output1 1 22 3 103 12 416024 question: (0, 0) --- (n, n) ask you how many paths there are; do not pass through the diagonal line. Train of Thought: separated by diagonal lines, triangle and triangle symmetric; reprinted please indicate the source: Looking for and star child question link: http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2067

 

 

 1 #include<stdio.h> 2 #define LL __int64 3 LL num[36][36]={0}; 4 void init() 5 { 6     for(int i=1;i<=35;i++) 7     { 8         num[i][0]=1; 9         for(int j=1;j<i;j++)10             num[i][j]=num[i][j-1]+num[i-1][j];11         num[i][i]=num[i][i-1];12     }13 }14 int main()15 {16     int n,ca=1;17     init();18     while(scanf("%d",&n)!=EOF)19     {20         if(n==-1) break;21         printf("%d %d %I64d\n",ca++,n,2*num[n][n]);22     }23     return 0;24 25 }

 

 

 

 

Add previous Code

#include <stdio.h> int main()  {      int i,j;      __int64 a[36] = {1};      __int64 b[36] = {0};      for (i=1;i<36;i++)      {          for(j=1;j<i;j++)              a[j]=a[j]+a[j-1];          b[i]=a[i]=a[i-1];      }        for(j=1;scanf("%d",&i),i;j++)    {        if(i==-1)            break;        else            printf("%d %d %I64d\n",j,i,2*b[i]);    }    return 0;  }  

I found that my previous questions are different in thinking...