链接
题解
一共往下走 次,其中 次往右走,所以答案是
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;
}