【题解】Luogu P5300 [GXOI/GZOI2019]与或和

原题传送门

我们珂以拆位，拆成一个个0/1矩阵

贡献珂以用全0，全1的子矩阵的个数来计算

全0，全1的子矩阵的个数珂以用悬线法/单调栈解决

#include <bits/stdc++.h>
#define N 1005
#define mod 1000000007
#define getchar nc
using namespace std;
inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
    register int x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
    return x*f;
}
inline void write(register int x)
{
    if(!x)putchar('0');if(x<0)x=-x,putchar('-');
    static int sta[20];register int tot=0;
    while(x)sta[tot++]=x%10,x/=10;
    while(tot)putchar(sta[--tot]+48);
}
int n,a[N][N],len[N],all,ans1,ans2; 
pair<int,int>s[N];
inline int work(register int t,register int opt)
{
	int res=0;
	for(register int i=1;i<=n;++i)
		len[i]=0;
	for(register int i=1;i<=n;++i)
	{
		int top=0,sum=0;
		for(register int j=1;j<=n;++j)
		{
			len[j]=(a[i][j]>>t&1)==opt?len[j]+1:0;
			int tmp=1;
			while(top&&s[top].first>=len[j])
				sum-=s[top].first*s[top].second,tmp+=s[top--].second;
			sum+=len[j]*tmp;
			res=(res+sum)%mod;
			s[++top]=make_pair(len[j],tmp);
		}
	}
	return res;
}
int main()
{
	n=read();
	for(register int i=1;i<=n;++i)
		for(register int j=1;j<=n;++j)
			a[i][j]=read(),all=(all+i*j)%mod;
	for(register int i=0,j=1;i<31;++i,j<<=1)
	{
		ans1=(ans1+1ll*work(i,1)*j)%mod;
		ans2=(ans2+1ll*(all-work(i,0)+mod)*j)%mod;
	}
	write(ans1),putchar(' '),write(ans2);
	return 0;
}