Cattle passenger 197C expect operand

Effect: Given $ x, q $, $ X $ each step becomes equal probability $ [x, q] $ any one number, find becomes a desired operand of $ Q $.

 

It is easy to get $ f (x, q) = \ frac {\ sum \ limits_ {i = x + 1} ^ qf (i, q) + qx + 1} {qx} $, $ F boundary conditions ( q, q) = 0 $.

Each query complexity is $ O (q) $, but the answer can be found only on the difference between the $ X $ and $ Q $, so the pre-treatment results to $ q = at N-1 $.

#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;
typedef long long ll;
const int P = 998244353;

const int N = 1e7+10;
int dp[N], inv[N];

int main() {
	inv[0]=inv[1]=1;
	REP(i,2,N-1) inv[i]=(ll)inv[P%i]*(P-P/i)%P;
	int sum = 0;
	PER(i,0,N-2) {
		dp[i] = (ll)(sum+N-i)*inv[N-1-i]%P;
		(sum += dp[i]) %= P;
	}
	int t;
	scanf("%d", &t);
	while (t--) {
		int x, q;
		scanf("%d%d", &x, &q);
		printf("%d\n",dp[N-1-q+x]);
	}
}