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。

 

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" alt="" />

 

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,+(n-m)*m);
     }
     return ;
 }