题目链接：https://www.nowcoder.com/acm/contest/140/J
思路：...可以随机权值减少碰撞，权值大的话，多个数加起来是一个是的倍数的可能性就比较小了。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <unordered_map>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e12+7)
#define pb push_back
#define lc (d<<1)
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int n,m,t,b[1000008];
ll a[1000008],qw;
vector<ll>g[1000008];
int main()
{
cin.tie(0);
cout.tie(0);
srand((int)time(0));
cin>>n>>m>>t;
FOR(i,0,n+1) g[i].resize(m+5);
FOR(i,1,n*m)
FOR(j,1,2) a[i]=a[i]+(ll)rand()*100000ll+(ll)rand();
FOR(i,1,n*m) si(b[i]);
int x,y,q,w,z,s=0;
while(t--)
{
si(x),si(y),si(q),si(w),si(z);
g[x][y]+=a[z];
g[x][w+1]-=a[z];
g[q+1][y]-=a[z];
g[q+1][w+1]+=a[z];
}
FOR(i,1,n*m)
{
x=i/m+1;
y=i%m;
if(y==0) y=m,x--;
qw=g[x][y]+=g[x-1][y]+g[x][y-1]-g[x-1][y-1];
if(qw%a[b[i]]) s++;
}
cout<<s<<endl;
return 0;
}