CodeForces 630G：Challenge Pennants【排列组合】

最新推荐文章于 2019-05-07 20:22:40 发布

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本文介绍了一个IT公司为了奖励员工而引入的一种非财务奖励系统——挑战旗帜，并探讨了如何将有限数量的不同类型旗帜分配到多个办公桌上，使得每面旗帜都位于一张桌子上，同时考虑到每个桌子可以放置任意数量的旗帜。

Challenge Pennants

Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

Because of budget cuts one IT company established new non-financial reward system instead of bonuses.

Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.

Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.

In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.

One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.

Output

Output one integer — the amount of ways to place the pennants on n tables.

Sample Input

Input

Output

所有的旗子都要用上，好吧，是我想多了~~~

AC-code：

#include<cstdio>
#include<cmath>
long long C(long long n,int x)
{
	long long temp=1;
	for(int i=1;i<=x;i++)
		temp=temp*(n-i+1)/i; 
	return temp;
}
int main()
{
	int n;
	long long a,b;
	scanf("%d",&n);
	a=C(n,5)+4*C(n,4)+6*C(n,3)+4*C(n,2)+C(n,1);
	b=C(n,3)+2*C(n,2)+C(n,1);
	printf("%lld\n",a*b);
	return 0;
}