GDUFE ACM-1020

The Rascal Triangle

Time Limit: 1000ms

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) the 

 element of the 

 row 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 



这道题就是找规律：
写成这样就可以发现规律了

 

 

AC代码：

#include<bits/stdc++.h>
using namespace std;

int main()
{
    long long  n,m,k,b,j,c,i,t,a[50005];
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld%lld%lld",&k,&n,&m);
        if(m==0||m==n)
            printf("%lld 1\n",k);
        else
        {
            b=n-3;
            a[0]=1;
            a[1]=n;
            for(i=2;b>0;i++)
            {
                a[i]=a[i-1]+b;
                b=b-2;
            }
            if(n%2==1)
            {
                if(m<=n/2)
                    printf("%lld %lld\n",k,a[m]);
                else
                {
                    j=(n+1)/2;
                    c=(m+1)%j;
                    m=j-c;//后半部分,比如n=7时,a[4]=a[3],a[5]=a[2],a[6]=a[1];
                    printf("%lld %lld\n",k,a[m]);
                }
            }
            else
            {
                if(m<=n/2)
                    printf("%lld %lld\n",k,a[m]);
                else
                {
                    j=n/2;
                    c=m%j;
                    m=j-c;//类似上面
                    printf("%lld %lld\n",k,a[m]);
                }
            }
        }
    }
    return 0;
}

 另附上高人的代码（我没有想到这个思路）：

#include <stdio.h>
//规律：（n-m）*m+1
int main(){
    int t,N,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d %d",&N,&n,&m);
        printf("%d %d\n",N,1+(n-m)*m);
    }
    return 0;
}