C++14(g++5.4) 解法, 执行用时: 1104ms, 内存消耗: 166764K, 提交时间: 2018-09-28 21:14:38

#include<bits/stdc++.h>
using namespace std;
const int mod=998244353;
long long sum[10000007],inv[10000007];
int main() {
    sum[0]=-1;
    sum[1]=1;
    inv[1]=1;
	for(int i=2;i<=10000000;++i) {
        inv[i]=1ll*(mod-(mod/i))*inv[mod%i]%mod;
        sum[i]=sum[i-1]+inv[i];
		if(sum[i]>=mod) sum[i]-=mod;
	}
	int T,x,q;
	scanf("%d",&T);
	while(T--) {
		scanf("%d%d",&x,&q);
		printf("%lld\n",sum[q-x]==mod-1?0:sum[q-x]+1);
	}
	return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 817ms, 内存消耗: 53220K, 提交时间: 2018-09-28 20:33:03

#include<cstdio>
const int p=998244353;
const int N=11000000;
int inv[N],T,x,q;
int main(){
	scanf("%d",&T);inv[1]=inv[0]=1;
	for (int i=2;i<N;i++) inv[i]=(long long)((long long)(p-p/i)*(long long)inv[p%i])%p;
	for (int i=1;i<N;i++) inv[i]=(inv[i]+inv[i-1])%p;
	while (T--){
		scanf("%d%d",&x,&q);
		printf("%d\n",inv[q-x]);
	}
	return 0;
}