【洛谷P5300】【GXOI/GZOI2019】—与或和（单调栈）

传送门

每一位$O(n^2)$做一下就完了

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>  
cs int mod=1e9+7;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=1005;
int n,a[N][N],f[N][N],up[N][N],ans1,ans2,tmp;
int q[N],top;
inline int calc(){
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)
		up[i][j]=(f[i][j]==0)?0:(up[i-1][j]+1);
	}
	int res=0;
	for(int i=1;i<=n;i++){
		int tot=0,top=0,*p=up[i];
		for(int j=1;j<=n;j++){
			while(top&&p[j]<=p[q[top]])Dec(tot,mul(dec(q[top],q[top-1]),p[q[top]])),top--;
			q[++top]=j,Add(tot,mul(dec(q[top],q[top-1]),p[j]));
			Add(res,tot);
		}
	}
	return res;
}
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	n=read();int tot=n*(n+1)/2;tot=mul(tot,tot);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n;j++)
	a[i][j]=read();
	for(int t=30;~t;t--){
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		f[i][j]=(bool)(a[i][j]&(1<<t));
		Add(ans1,mul(calc(),(1<<t)%mod));
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
		f[i][j]^=1;
		Add(ans2,mul(dec(tot,calc()),(1<<t)%mod));
	}
	cout<<ans1<<" "<<ans2<<'\n';
}