Wannafly挑战赛25C 期望操作数

Wannafly挑战赛25C 期望操作数

---

简单题啦

(f[i]=frac{sum_{j<=i}f[j]}{i}+1)

(f[i]=frac{f[i]}{i}+frac{sum_{j<i}f[j]}{i}+1)

(frac{i-1}{i}f[i]=frac{sum_{j<i}f[j]+i}{i})

(f[i]=frac{sum_{j<i}f[j]+i}{i-1})

一边求逆元一边dp即可

#include<bits/stdc++.h>
#define il inline
#define vd void
#define mod 998244353
typedef long long ll;
il int gi(){
	int x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch)){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return x*f;
}
ll f[10000001],inv[10000001];
int T,l,r;
int main(){
#ifndef ONLINE_JUDGE
	freopen("197c.in","r",stdin);
	freopen("197c.out","w",stdout);
#endif
	f[0]=0;f[1]=2;ll S=0;inv[1]=1;
	for(int i=2;i<10000001;++i)f[i]=(1+(S+1)*inv[i-1])%mod,S=(S+f[i])%mod,inv[i]=(mod-(mod/i*inv[mod%i])%mod)%mod;
	T=gi();
	while(T--)l=gi(),r=gi(),printf("%lld
",f[r-l+1]);
	return 0;
}