nowcoder 181045 / 克洛涅的多项式 构造+思维

code:

#include <bits/stdc++.h>
#define ll long long
#define mod 998244353
#define N 1000006
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
namespace IO
{
	char *p1, *p2, buf[100000];
	#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )
	int rd() { int x = 0;char c = nc();while (c < 48) {c = nc();}while (c > 47) {x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();}return x;}
};
ll qpow(ll base,ll k)
{
	ll tmp=1;
	for(;k;k>>=1,base=base*base%mod) if(k&1) tmp=tmp*base%mod;
	return tmp;
}
int a[N];
int main()
{
	using namespace IO;
	// setIO("input");
	int n=rd(),m=rd(),k=rd(),i,j,ans=1;
	for(i=1;i<=n;++i)
	{
		int x=rd();
		ans=(ll)ans*(k-x)%mod;
	}
	int mdl=1;
	for(i=0;i<=m;++i)
	{
		int x=rd();
		(ans+=(ll)mdl*x%mod)%=mod;
		mdl=(ll)mdl*k%mod;
	}
	printf("%d\n",ans);
	return 0;
}