#include <bits/stdc++.h>
using namespace std;
int g[1001][1001], c[1001], r[1001]; //g[r][c]
int h, w, ans;
long long work(int ans)
{
long long p=1;
for (int q=1;q<=ans;q++)
{
p = (p * 2) % (1000000007);
}
return p;
}
int main()
{
memset(g, 0, sizeof(g));
cin >> h >> w;
for (int i = 1; i <= h; i++)
{
cin >> r[i];
for (int j = 1; j <= r[i]; j++)
{
if (g[i][j] == 2)
{
cout << 0;
return 0;
}
g[i][j] = 1;
}
if (r[i] < w && g[i][r[i] + 1] == 1)
{
cout << 0;
return 0;
}
g[i][r[i] + 1] = 2;
}
for (int i = 1; i <= w; i++)
{
cin >> c[i];
for (int j = 1; j <= c[i]; j++)
{
if (g[j][i] == 2)
{
cout << 0;
return 0;
}
g[j][i] = 1;
}
if (c[i] < h && g[c[i] + 1][i] == 1)
{
cout << 0;
return 0;
}
g[c[i] + 1][i] = 2;
}
for (int i = 1; i <= h; i++)
for (int j = 1; j <= w; j++)
if (g[i][j]==0) ans++;
cout << work (years);
return 0;
}
Black white black and white Each row can not be changed determine if there is a direct conflict to return.
First memset0, black and white is 2 to 1, and if there is a conflict in the square directly cout << - 1 then halt;
Finally, statistics do not need points, permutations (this question persecution junior high school students) Cn, 0 + Cn, 1 + `` `` `` + Cn, n = 2 ^ n
That operator would have a 2 ^ n
Why do I so slow? !
Not initially understand the topic, not black and white -1 consideration.
Followed by the keyboard sucks my shift shells do not press down to buckle up have their own
Hrwcij already ranks around dizzy.
Now I have decided, in line with debug, all rows in all on the first [line], all on the second row.
Regardless of pit father entitled to a table on which quadrant, I am his mother so treated. I changed Ij for a long time.
PS This question seems to have a lot of people comment when heavy kneeling, wait a minute I looked at the code pull a -1