CodeForces - 509A Maximum in Table （类杨辉三角，构建数组）

最新推荐文章于 2019-03-03 09:19:37 发布

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最新推荐文章于 2019-03-03 09:19:37 发布

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探讨了类杨辉三角的二维数组构造方法及其最大值的求解过程，通过递推公式Ai,j=Ai-1,j+Ai,j-1填充数组，并输出n×n数组中的最大元素。

An n × n table a is defined as follows:

The first row and the first column contain ones, that is: ai, 1 = a1, i = 1 for all i = 1, 2, …, n.
Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai, j = ai - 1, j + ai, j - 1.
These conditions define all the values in the table.

You are given a number n. You need to determine the maximum value in the n × n table defined by the rules above.

Input
The only line of input contains a positive integer n (1 ≤ n ≤ 10) — the number of rows and columns of the table.

Output
Print a single line containing a positive integer m — the maximum value in the table.

Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:

{1, 1, 1, 1, 1},
{1, 2, 3, 4, 5},
{1, 3, 6, 10, 15},
{1, 4, 10, 20, 35},
{1, 5, 15, 35, 70}.
问题链接：http://codeforces.com/problemset/problem/509/A
问题简述：类杨辉三角的二维数组，输入n，求数组中的最大值
问题分析：题目中已给出Ai,j=Ai-1,j+Ai,j-1。把Ai,0，A0,j全部赋值0，再用前面的公式可以求出数组，再输出A[n-1][n-1]即可。
AC通过的C++语言程序如下：

#include<iostream>
#include <algorithm>
#include<iostream>
#include<string>
#include<stdio.h>
#include <algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
	int n;
	cin >> n;
	int **p = new int*[n];
	for (int i = 0; i < n; i++)
	{
		p[i] = new int[n];
	}
	p[0][0] = 1;
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			p[i][0] = 1;
			p[0][j] = 1;
			if (i > 0 && j > 0) p[i][j] = p[i - 1][j] + p[i][j - 1];
		}
	}
	cout << p[n - 1][n - 1];
	return 0;
}