洛谷 P6835 [Cnoi2020]线形生物 数学期望+递推

链接
根据期望定义得出相邻两点步数的期望，利用前缀和，期望的线性性

质进行优化即可，式子就不写了(不会用latex)注意，所得的

期望因取模，可能变成负值，加个模数即可(卡了我10分钟)

#include<cstring>
#include<string>
#include<cstdio>
#define ll long long
#define mod 998244353
using namespace std;
const int maxn=2000005;
inline ll read(){
	ll ret=0;
	ll f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')
			f=-f;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		ret=ret*10+(ch^'0');
		ch=getchar();
	}
	return ret*f;
}
ll id,n,m;
struct edge{
	ll nex;
	ll to;
}e[maxn];
ll head[maxn];
ll cnt;
void add(ll u,ll v){
	cnt++;
	e[cnt].nex=head[u];
	e[cnt].to=v;
	head[u]=cnt;
}
ll f[maxn];
ll u,v;
ll sum[maxn];
ll fz[maxn];
int main(){
//	freopen("a.in","r",stdin);
	id=read();
	n=read();
//	cin>>n>>m;
	m=read();
	for(int i=1;i<=m;i++){
	//	cin>>u>>v;
		u=read();
		v=read();
		add(u,v);
		fz[u]++;
	}
	for(int i=1;i<=n;i++){
		f[i]=fz[i]+1;
		for(int j=head[i];j;j=e[j].nex){
			int y=e[j].to;
			f[i]=((f[i]+(sum[i-1]-sum[y-1])%mod)%mod+mod);
		}
		sum[i]=(sum[i-1]+f[i])%mod;
	}
	cout<<sum[n]<<endl;
	return 0;
}