HDU 4237 The Rascal Triangle

The Rascal Triangle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 243    Accepted Submission(s): 192

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.

 

 

Sample Input

5

1 4 0

2 4 2

3 45678 12345

4 12345 9876

5 34567 11398

 

 

Sample Output

1 1

2 5

3 411495886

4 24383845

5 264080263

 

 

Source

2011 Greater New York Regional

 

 

 

解析：利用题中所给递推公式，将数据存放到数组中有时候是个不错的选择。但本题这样做很可能会MLE，所以应该考虑找出通项公式。可以先写个小程序输出The Rascal Triangle的前几行，观察找规律。

 

 

很容易发现通项公式为1+(n-m)*m。

 

 

 

#include <cstdio>

int P;
int N,n,m;

int main()
{
    scanf("%d",&P);
    while(P--){
        scanf("%d%d%d",&N,&n,&m);
        printf("%d %d\n",N,1+(n-m)*m);
    }
    return 0;
}