題意:
題解:
AC代碼
/*
Author:zzugzx
Lang:C++
Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define SZ(x) (int)x.size()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e5+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
const int p=1e8;
int n,m;
ll g[20][20],dp[20][5000];
bool ok(int st,int i){
for(int j=0;j<m;j++)
if((st>>j)&1&&!g[i][j])return 0;
return 1;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(cin>>n>>m){
memset(dp,0,sizeof dp);
for(int i=1;i<=n;i++)
for(int j=0;j<m;j++)
cin>>g[i][j];
dp[0][0]=1;
for(int i=1;i<=n;i++){
for(int st=0;st<(1<<m);st++){
if(st&(st>>1)||!ok(st,i))continue;
for(int st1=0;st1<(1<<m);st1++){
if(st1&st)continue;
dp[i][st]=(dp[i][st]+dp[i-1][st1])%p;
}
}
}
ll ans=0;
for(int i=0;i<(1<<m);i++)ans=(ans+dp[n][i])%p;
cout<<ans<<endl;
}
return 0;
}