J Less taolu

链接：https://ac.nowcoder.com/acm/contest/338/J
来源：牛客网

题目描述

Less taolu, more sincerity.

This problem is very easy to solve.

You may be very tired during this contest. So we prepared a gift for you.

You just copy and paste this code and you will get AC!

Ctrl + C && Ctrl + V is a necessary skill for a programming ape.

#include<iostream>
using namespace std;
const long long mod = 1e9+7;
long long func(int x){
    if (x==1||x==0){
        return 1;
    }
    return (x*func(x-1)+(x-1)*func(x-2))%mod;
}
int n;
int main(){
    cin>>n;
    cout<<func(n);
    return 0;
}

 

 

输入描述:

Input only a single
integer n.

输出描述:

Please output
the answer by this code.

示例1

输入

复制

3

输出

复制

11

示例2

输入

复制

100

输出

复制

372497045

备注:

0<=N<=1e5

记忆化搜索，懒得解释

#include<iostream>
using namespace std;
const int maxn = 100000+10;
int f[maxn];
const long long mod = 1e9+7;
long long func(int x)
{
    if (x==1||x==0)
    {
        return 1;
    }
    if(f[x])
        return f[x];
    else
        return f[x] = (x*func(x-1)+(x-1)*func(x-2))%mod;
}
int n;
int main()
{
    cin>>n;
    cout<<func(n);
    return 0;
}