题目链接：https://codeforc.es/contest/1452/problem/D
There are n+2 towns located on a coordinate line, numbered from 0 to n+1. The i-th town is located at the point i.
You build a radio tower in each of the towns 1,2,…,n with probability 12 (these events are independent). After that, you want to set the signal power on each tower to some integer from 1 to n (signal powers are not necessarily the same, but also not necessarily different). The signal from a tower located in a town i with signal power p reaches every city c such that |c?i|<p.
After building the towers, you want to choose signal powers in such a way that:
towns 0 and n+1 don’t get any signal from the radio towers;
towns 1,2,…,n get signal from exactly one radio tower each.
For example, if n=5, and you have built the towers in towns 2, 4 and 5, you may set the signal power of the tower in town 2 to 2, and the signal power of the towers in towns 4 and 5 to 1. That way, towns 0 and n+1 don’t get the signal from any tower, towns 1, 2 and 3 get the signal from the tower in town 2, town 4 gets the signal from the tower in town 4, and town 5 gets the signal from the tower in town 5.
Calculate the probability that, after building the towers, you will have a way to set signal powers to meet all constraints.
Input
The first (and only) line of the input contains one integer n (1≤n≤2?105).
Output
Print one integer — the probability that there will be a way to set signal powers so that all constraints are met, taken modulo 998244353.
Formally, the probability can be expressed as an irreducible fraction xy. You have to print the value of x?y?1mod998244353, where y?1 is an integer such that y?y?1mod998244353=1.
Examples
input
2
output
748683265
input
3
output
748683265
input
5
output
842268673
input
200000
output
202370013
分析
n = 1 时，只有一种放法
n = 2 时，也只有一种方法
n = 3 时，有两种放法
n = 4时，有三种
n = 5时，有五种
……
我们发现分子就是斐波那契数列，分母则为 2 的 n 次方。
代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;const ll mod = 998244353;
ll f[200007];
ll n;ll qm(ll a,ll b)
{
ll res = 1;while(b){
if(b & 1) res = (res * a) % mod;a = (a * a) % mod;b >>= 1;}return res % mod;
}int main()
{
f[1] = f[2] = 1;for(int i=3;i<=200000;i++) f[i] = (f[i - 1] + f[i - 2]) % mod;scanf("%lld",&n);ll ans = f[n] * qm(qm(2, n), mod - 2) % mod;printf("%lld",ans);return 0;
}