牛客网第二场Jfarm（随机化+二维前缀和）

链接：https://www.nowcoder.com/acm/contest/140/J
来源：牛客网

White Rabbit has a rectangular farmland of n*m. In each of the grid there is a kind of plant. The plant in the j-th column of the i-th row belongs the a[i][j]-th type.
White Cloud wants to help White Rabbit fertilize plants, but the i-th plant can only adapt to the i-th fertilizer. If the j-th fertilizer is applied to the i-th plant (i!=j), the plant will immediately die.
Now White Cloud plans to apply fertilizers T times. In the i-th plan, White Cloud will use k[i]-th fertilizer to fertilize all the plants in a rectangle [x1[i]...x2[i]][y1[i]...y2[i]].
White rabbits wants to know how many plants would eventually die if they were to be fertilized according to the expected schedule of White Cloud.


输入描述:
The first line of input contains 3 integers n,m,T(n*m<=1000000,T<=1000000)
For the next n lines, each line contains m integers in range[1,n*m] denoting the type of plant in each grid.
For the next T lines, the i-th line contains 5 integers x1,y1,x2,y2,k(1<=x1<=x2<=n,1<=y1<=y2<=m,1<=k<=n*m)
输出描述:
Print an integer, denoting the number of plants which would die.
示例1
输入

复制
2 2 2
1 2
2 3
1 1 2 2 2
2 1 2 1 1
输出

复制
3

随机化，避免4=1+3=2+2

让他只有2+2；

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<vector>
using namespace std;
#define ll long long
int n,m,t;
int b[1000005];
ll a[1000008],qw;
vector<ll>g[1000008];
int x,y,q,w,z;
int main()
{
    srand(time(0));
    scanf("%d%d%d",&n,&m,&t);
    for(int i=0;i<=n+1;i++)
        g[i].resize(m+5);
    for(int i=0;i<=n*m;i++)
    {
        a[i]=(ll)rand()*1000001+(ll)rand();
        a[i]=a[i]+(ll)rand()*1000001+(ll)rand();
    }
    for(int i=1;i<=n*m;i++)
        scanf("%d",&b[i]);
    while(t--)
    {
        
        scanf("%d%d%d%d%d",&x,&y,&q,&w,&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];
        
    }
    int ans=0;
    for(int i=1;i<=n*m;i++)
    {
        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]]) ans++;
    }
    printf("%d",ans);
    
    return 0;
}