Blue Bridge Cup training solution to a problem - hunting Matrix (issue 1097)

Original title link: http: //www.dotcpp.com/oj/problem1097.html

Title Description

Serpentine matrix is ​​a triangular matrix of a natural number 1 sequentially arranged in the beginning.

Entry

The title plurality of sets of data, each data consisting of integer N by one. (N is not larger than 100)

Export

For each set of data and outputs a matrix of N rows of serpentine. Do extra blank line between the two sets of output. Digital Matrix triangle in the same line separated by a space. End of the line do extra spaces.

Sample input

5

Sample Output

1 3 6 10 15
2 5 9 14
4 8 13
7 12
11

Problem-solving ideas

0 - 15. 1. 3. 6 10
12 is 14. 9. 5
2 - 13 is. 4. 8
. 3 - 12 is. 7
. 4 -. 11
us analyze, problems like this are to be linked to the rows and columns,
for the first row 0 The number is 1 1 + 2 (0 + 2) 1 + 2 + 3 1 + 2 + 3 + 4 1 + 2 + 3 + 4 + 5
for the first number is the first line 2 2 + 3 (1 + 2 ) 2 + 3 + 4 2 + 3 + 4 + 5
for the second row is the first number 4 4 + 4 (2 + 2) 4 + 4 + 5
for the third row first number is 7 . 7 + 5 ( 3 + 2)
to the first number is the fourth row 11
by observing the second number per row = first row i + 2 + number is followed by the accumulation of long plus q determines the starting only thereafter q is incremented by summing need to be the i-th row and q = +2
had then determine where the start number in the line
0 -. 1
. 1 - 2
2 -. 4
. 3 -. 7
. 4 -11
Where it is easy to see (the start line number 0 is the set of 1) (starting point of the previous line) 2 = 1 (starting value of the previous line) + 1 (current number of lines) = 4 2 + 2 ( the current line number) 7 = 4 (on line starting value) + 3 (the current line number) 11 = 7 (on line starting value) +4 (the current number of rows)
so what we are dealing with an array of
written expression: current initial value equal to the current line number of a line +
expression: a [i] = a [ i-1] + i
Another point is that
from the beginning of the first row, the number of elements on each row line less than a

Reference Code

#include<iostream>
#include<cstdio>
#define Max 101
using namespace std;
int main()
{
	int n,a[Max];
	int p=1,q;
	int k;
	//循环录入每行的初始值 
	for(int i = 0;i < 101;++i)
	{
		if(i != 0)
		{
			a[i] = a[i-1]+i;
		}
		else
		{
			a[i] = 1;
		}
		
	}
	while(cin>>n)
	{
		//k是列数   
		k = n;
		for(int i=0;i <n;++i)
		{
			 
			p = a[i];
			q = i+2;
			
			for(int j=0;j < k;++j)
			{
				if(j == 0)
				{
					cout << p <<" "; 
				}else
				{
					p=p+q;
					q +=1;
					//到达尾部就不要输出空格 
					if(j ==k-1)
					{
						cout << p ;
					}else
					{
						cout << p <<" ";
					}
				}
			}
			--k; //列数相对上一行少1 
			cout<<endl;
			
		}
	}
	return 0;
}