cf509A Maximum in Table

A. Maximum in Table

time limit per test 2 seconds

memory limit per test 256 megabytes

input standard input

output standard output

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.

Sample test(s)

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}.

这题不需要解释了吧

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<ctime>
#include<iomanip>
#define LL long long
#define inf 0x7ffffff
#define N 200010
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
LL a[11][11];
int n;
int main()
{
	n=read();
	a[0][1]=1;
	for (int i=1;i<=10;i++)
		for (int j=1;j<=10;j++)
			a[i][j]=a[i-1][j]+a[i][j-1];
	printf("%lld
",a[n][n]);
}