1 #include<cstdio>
 2 typedef long long ll;
 3 const int N=3233;
 4 int a[N][N],sta[N],maxh[N];
 5 ll res[N];
 6 int main()
 7 {
 8     int n,m;
 9     scanf("%d%d",&n,&m);
10     for(int i=1;i<=n;i++)
11         for(int j=1;j<=m;j++)
12             scanf("%d",&a[i][j]);
13     int top;
14     ll all=(1ll*n*(n+1)*m*(m+1))>>2,ans=0ll;//all总的方案数
15     for(int i=1;i<=n;i++)
16     {
17         sta[top=0]=0;
18         for(int j=1;j<=m;j++)
19         {
20             if(!a[i][j])//因为我们要求的是全1,所以维护这一列0的最大行数，也就是高度
21                 maxh[j]=i;
22             while(top&&maxh[sta[top]]<=maxh[j])
23                 top--;
24             res[j]=res[sta[top]]+1ll*(j-sta[top])*maxh[j];
25             ans+=res[j];
26             sta[++top]=j;
27         }
28     }
29     printf("%lld\n",all-ans);
30     return 0;
31 }

 1 #include<cstdio>
 2 typedef long long ll;
 3 const int N=1108,mod=1000000007;
 4 int n,cf2[32]={1},a[N][N],h[N][N];
 5 int sta[N],maxh[N];
 6 ll res[N];
 7 ll count(int k,int x)
 8 {
 9     for(int i=1;i<=n;i++)
10         maxh[i]=0;
11     int top=0;
12     ll ans=0;
13     for(int i=1;i<=n;i++)
14     {
15         sta[top=0]=0;
16         for(int j=1;j<=n;j++)
17         {
18             if(a[i][j]&cf2[k])
19                 h[i][j]=x;
20             else
21                 h[i][j]=x^1;
22             if(!h[i][j])
23                 maxh[j]=i;
24             while(top&&maxh[sta[top]]<=maxh[j])
25                 top--;
26             res[j]=res[sta[top]]+1ll*(j-sta[top])*maxh[j];
27             if(res[j]>=mod)
28                 res[j]-=mod;
29             ans+=res[j];
30             if(ans>=mod)
31                 ans-=mod;
32             sta[++top]=j;
33         }
34     }
35     return ans;
36 }
37 int main()
38 {
39     for(int i=1;i<=30;i++)
40         cf2[i]=cf2[i-1]<<1;
41     scanf("%d",&n);
42     for(int i=1;i<=n;i++)
43         for(int j=1;j<=n;j++)
44             scanf("%d",&a[i][j]);
45     ll all=((1ll*n*(n+1)*n*(n+1))>>2)%mod;
46     ll ans1=0,ans2=0;
47     for(int k=0;k<=30;k++)
48     {
49         ans1+=((all-count(k,1)+mod)%mod)*1ll*cf2[k];
50         if(ans1>=mod)
51             ans1%=mod;
52         ans2+=count(k,0)*1ll*cf2[k];
53         if(ans2>=mod)
54             ans2%=mod;
55     }
56     printf("%lld %lld\n",ans1,ans2);
57     return 0;
58 }