HDU 6397 Character Encoding

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HDU 6397 Character Encoding [ 容斥原理 + 组合数 ]
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=998244353; 
const int N=2e5+10;
ll inv[N],pre[N];
ll qpow(ll x,ll n){
	ll res=1;
	while(n){
		if(n%2==1) res=res*x%mod;
		x=x*x;x%=mod;
		n=n>>1;
	}
	return res;
}
void init(){
	pre[0]=inv[0]=1;
	for(int i=1;i<N;++i){pre[i]=pre[i-1]*i;pre[i]%=mod;}
	inv[N]=qpow(pre[N],mod-2);
	for(int i=N-1;i>=1;--i){inv[i]=inv[i+1]*(i+1);inv[i]%=mod;}
}
ll C(int n,int m){
	if(m<0||m>n)return 0;
	return pre[n]*inv[n-m]%mod*inv[m]%mod;
}
int t,n,m,k;
ll ans;
int main(){
	init();
	scanf("%d",&t);
	while(t--){
		ans=0;
		scanf("%d%d%d",&n,&m,&k);
		for(int i=0;i<=min(k/n,m);++i){
			if(i&1)ans=(ans-C(m,i)*C(k-i*n+m-1,m-1)%mod+mod)%mod;
			else ans=(ans+C(m,i)*C(k-i*n+m-1,m-1)%mod+mod)%mod;
		}
		printf("%lld\n",ans);
	}
	return 0;
}