LOJ#3083. 「GXOI / GZOI2019」与或和
显然是先拆位，AND的答案是所有数字为1的子矩阵的个数
OR是所有的子矩阵个数减去所有数字为0的子矩阵的个数
子矩阵怎么求可以记录每个位置能向上延伸的高度\(h[i][j]\)
枚举左下角的端点，用一个单调栈维护即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,all;
int a[1005][1005],b[1005][1005],ans[2],h[1005],sta[1005],top;
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void update(int &a,int b) {
a = inc(a,b);
}
int getall(int x) {
memset(h,0,sizeof(h));
int res = 0,sum = 0;
for(int i = 1 ; i <= N ; ++i) {
sum = 0;top = 0;sta[0] = N + 1;
for(int j = N ; j >= 1 ; --j) {
if(b[i][j] == x) h[j]++;
else h[j] = 0;
while(top && h[sta[top]] >= h[j]) {
update(sum,MOD - mul(sta[top - 1] - sta[top],h[sta[top]]));
--top;
}
sta[++top] = j;
update(sum,mul(sta[top - 1] - sta[top],h[j]));
update(res,sum);
}
}
return res;
}
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {
for(int j = 1 ; j <= N ; ++j) {
read(a[i][j]);
update(all,mul(N - i + 1,N - j + 1));
}
}
for(int i = 30 ; i >= 0 ; --i) {
for(int j = 1 ; j <= N ; ++j) {
for(int h = 1 ; h <= N ; ++h) {
b[j][h] = a[j][h] >> i & 1;
}
}
int t = (1 << i) % MOD;
update(ans[0],mul(t,getall(1)));
update(ans[1],mul(t,inc(all,MOD - getall(0))));
}
out(ans[0]);space;out(ans[1]);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}