hdu4237 The Rascal Triangle 规律题

The Rascal Triangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 64    Accepted Submission(s): 61

Problem Description

The Rascal Triangle definition is similar to that of the Pascal Triangle. The rows are numbered from the top starting with 0. Each row n contains n+1 numbers indexed from 0 to n. Using R(n,m) to indicte the index m item in the index n row:
   R(n,m) = 0 for n < 0 OR m < 0 OR m > n
The first and last numbers in each row(which are the same in the top row) are 1:
   R(n,0) = R(n,n) = 1
The interior values are determined by (UpLeftEntry*UpRightEntry+1)/UpEntry(see the parallelogram in the array below):
   R(n+1, m+1) = (R(n,m) * R(n,m+1) + 1)/R(n-1,m)

Write a program which computes R(n,m) theelement of therow of the Rascal Triangle.

 

Input

The first line of input contains a single integer P, (1 <= P <= 1000), which is the number of data sets that follow. Each data set is a single line of input consisting of 3 space separated decimal integers. The first integer is data set number, N. The second integer is row number n, and the third integer is the index m within the row of the entry for which you are to find R(n,m) the Rascal Triangle entry (0 <= m <= n <= 50,000).

 

Output

For each data set there is onr line of output. It contains the data set number, N, followed by a single space which is then followed by thr Rascal Triangle entry R(n,m) accurate to the integer value.

 

5
1 4 0
2 4 2
3 45678 12345
4 12345 9876
5 34567 11398

 

 输出

5
1 4 0
2 4 2
3 45678 12345
4 12345 9876
5 34567 11398

 

题意

R(n+1, m+1) = (R(n,m) * R(n,m+1) + 1)/R(n-1,m)

R(n,0) = R(n,n) = 1

数塔从第0层开始  每层也是从0 开始

根据上面的2个式子推出来所有的

问输入 a b  输出第a行第b个数字

 

    0                                        1

    1                                     1     1

    2                                  1     2       1

    3                               1     3     3        1

    4                            1     4     5       4        1

    5                        1      5     7     7        5        1

    6                    1      6      9     10      9        6      1

 

 

 

可以看出第i行第j个满足 （i-j）*j+1;

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哎 只有努力了   抓紧时间搞下规律题

#include<stdio.h>
int main()
{
 int a,i,j,T,k;
 scanf("%d",&T);
 while(T--)
 {
  scanf("%d %d %d",&a,&i,&j);
  if(i==j||j==0)
   printf("%d %d\n",a,1);
        else 
        {            
   k= (i-j)*j+1;          
   printf("%d %d\n", a, k); 
        } 
    } 
    return 0; 
}