BZOJ3326: [Scoi2013]数数

#include<cstdio>
typedef long long ll;
const int p=20130427;
const int N=1e5+5;
int n1,n2;
ll s1,z1[N],z2[N],c1[N],c2[N],c3[N],c4[N];
ll cal(ll s){return(s-1)*s/2%p;}
ll cal(int i,int j){
	return((j+1)*c3[i]-c4[i]+p)%p;
}
int sol(ll*f,int n){
	ll s2=0,s3=(c2[n]*cal(f[n])%p*c1[n]+cal(n-1,n)*cal(s1)%p*(f[n]-1)%p*c1[n-1])%p;
	for(int i=n-1;i;--i)
		s2=(s2+(n-i)*c2[i+1]%p*f[i+1])%p,s3=(s3+s2*f[i]%p*c1[i]+(n-i+1)*c2[i]%p*cal(f[i])%p*c1[i]+cal(i-1,n)*cal(s1)%p*f[i]%p*c1[i-1]+c2[i]*cal(s1)%p*c1[i]+cal(i-1,i)*cal(s1)%p*(s1-1)%p*c1[i-1])%p;
	return s3;
}
int main(){
	scanf("%lld%d",&s1,&n1);
	for(int i=n1;i;--i)
		scanf("%lld",z1+i);
	scanf("%d",&n2);
	for(int i=n2;i;--i)
		scanf("%lld",z2+i);
	int n3=n1<n2?n2:n1;
	c1[1]=1;
	for(int i=2;i<=n3;++i)
		c1[i]=s1*c1[i-1]%p;
	for(int i=1;i<=n3;++i)
		c2[i]=(c2[i-1]+c1[i])%p,c3[i]=(c3[i-1]+c2[i])%p,c4[i]=(c4[i-1]+c2[i]*i)%p;
	++z2[1];
	printf("%d\n",(sol(z2,n2)-sol(z1,n1)+p)%p);
}