CodeForces 509A Maximum in Table

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本文介绍了一个关于矩阵的问题，初始状态为第一行第一列都为1，其余元素为上一个元素与左侧元素之和，目的是找出该矩阵中的最大值。

Maximum in Table

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 509A

Description

An n × n table a is defined as follows:

The first row and the first column contain ones, that is: a_i, 1 = a_1, 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 a_i, j = a_{i - 1, j} + a_{i, 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.

Sample Input

Input

Output

Input

Output

Hint

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}.

题意：矩阵，开始时，第一行第一列的都是1，其余每一个数都是上边的数字和左边的数字之和，输出矩阵中最大的数。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
using namespace std;
int a[10][10];

int main(){
	int n;
	while(~scanf("%d", &n)){
		for(int i = 0; i < n; ++i){
			a[i][0] = a[0][i] = 1;
		}
		for(int i = 1; i < n; ++i){
			for(int j = 1; j < n; ++j){
				a[i][j] = a[i - 1][j] + a[i][j - 1];
			}
		}
		cout << a[n -1][n - 1] << endl; 
	}
}