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): 64Accepted 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

212 1

3 1 3 3 1

4 14 54 1

5 1 5 77 5 1

61 6 9 10 96 1

可以看出第i行第j个满足 （i-j）*j+1;

规律找了好久 最终还是看人家的 看着我的队友10几分钟就看出来了 我真的感到好丢人啊

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;
}