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题解

一共往下走$n-1$次，其中$m-1$次往右走，所以答案是$C_{n-1}^{m-1}{a}^{(n-1)-(m-1)}{b}^{m-1}$

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{ll c, f(1);for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;for(;isdigit(c);c=getchar())x=x*10+c-0x30;return f*x;
}
struct EasyMath
{ll prime[maxn], phi[maxn], mu[maxn];bool mark[maxn];ll fastpow(ll a, ll b, ll c){ll t(a%c), ans(1ll);for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;return ans;}void shai(ll N){ll i, j;for(i=2;i<=N;i++)mark[i]=false;*prime=0;phi[1]=mu[1]=1;for(i=2;i<=N;i++){if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;for(j=1;j<=*prime and i*prime[j]<=N;j++){mark[i*prime[j]]=true;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}mu[i*prime[j]]=-mu[i];phi[i*prime[j]]=phi[i]*(prime[j]-1);}}}ll inv(ll x, ll p)  //p是素数{return fastpow(x%p,p-2,p);}
}em;
ll fact[maxn], _fact[maxn];
#define mod 998244353ll
ll C(ll n, ll m)
{return fact[n] * _fact[m] %mod * _fact[n-m] %mod;
}
int main()
{ll i, ans, T=read(), a, b, n, m;fact[0]=_fact[0]=1;rep(i,2e5)fact[i]=fact[i-1]*i%mod, _fact[i]=em.inv(fact[i],mod);while(T--){a=read(), b=read(), n=read(), m=read();ans = em.fastpow(b,m-1,mod) * em.fastpow(a,n-m,mod) %mod * C(n-1,m-1) %mod;printf("%lld\n",ans);}return 0;
}