​​http://codeforces.com/contest/1006/problem/F​​

爆搜复杂度2^(n+m) n和m都不大 可以考虑折半搜索 即从起点终点各走一半

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pil pair <int,ll>

map <pil,ll> mp;
ll mat[50][50],prex[50],prey[50];
ll ans,k;
int n,m;

void dfsI(int x,int y,int step,ll val)
{
    pil tmp;
    if(x+y-2==step)
    {
        tmp=make_pair((x-1)*m+y,val);
        mp[tmp]++;
        return;
    }
    if(y+1<=m) dfsI(x,y+1,step,val^mat[x][y+1]);
    if(x+1<=n) dfsI(x+1,y,step,val^mat[x+1][y]);
}

void dfsII(int x,int y,int step,ll val)
{
    pil tmp;
    if(n-x+m-y==step)
    {
        tmp=make_pair((x-1)*m+y,k^val^mat[x][y]);
        ans+=mp[tmp];
        return;
    }
    if(y-1>=1) dfsII(x,y-1,step,val^mat[x][y-1]);
    if(x-1>=1) dfsII(x-1,y,step,val^mat[x-1][y]);
}

int main()
{
    int i,j,mid;
    scanf("%d%d%lld",&n,&m,&k);
    for(i=1;i<=n;i++) for(j=1;j<=m;j++) scanf("%lld",&mat[i][j]);
    mid=(n+m-2)/2;
    dfsI(1,1,mid,mat[1][1]);
    dfsII(n,m,(n+m-2)-mid,mat[n][m]);
    printf("%lld\n",ans);
    return 0;
}