HDU_4237_The Rascal Triangle(推公式)

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The Rascal Triangle

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

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 indicate 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)

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Write a program which computes R(n, m) the m-th element of the n-th row of the Rascal Triangle.

Input

The first line of input contains a single integer P, ( 1 $\le$ P $\le$ 1000), which is the number of data sets that follow. Each data set is a single line of input consisting of 3 spaces separated decimal integers. The first integer is data set number, N. The second integer is row numbern, 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 $\le$ m $\le$ n $\le$ 50000).

Output

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

Sample Input

51 4 02 4 23 45678 123454 12345 98765 34567 11398

Sample Output

1 12 53 4114958864 243838455 264080263

题型：找规律

题意：给出一个类似于杨辉三角的数塔，最左端和最右端都是1，递推式为

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

就是这个数等于其上面两个数的积再加1再除以其上面的上面的那个数，for example ：（3*3+1）/ 2 = 5。

求出第n行第m个数是什么。

分析：根据递推式来写代码是不能的，会超时。直接求出一个式子直接用就可以了，代码及其简短。

代码：

#include<iostream>#include<cstdio>using namespace std;int main(){int t,r;long long n,m,k;scanf("%d",&t);while(t--){scanf("%d%lld%lld",&r,&k,&m);printf("%d %lld\n",r,1+(k-m)*m);}return 0;}