Cube

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1949    Accepted Submission(s): 1013

Problem Description

Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N).
We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).
0: “Query” operation we want to get the value of A[i, j, k].

Input

Multi-cases.
First line contains N and M, M lines follow indicating the operation below.
Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation.
If X is 1, following x1, y1, z1, x2, y2, z2.
If X is 0, following x, y, z.

Output

For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)

Sample Input

2 5
1 1 1 1 1 1 1
0 1 1 1
1 1 1 1 2 2 2
0 1 1 1
0 2 2 2

Sample Output

1
0
1

Author

alpc32

Source

2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

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/*
题意实在太模糊了，两种操作0：单点查询，这个点的数字是多少
1：将某一区间内的数字都反转；
只需要维护反转次数的前缀和就行了

编译器出毛病了，宏定义不行，调用函数也不行
*/

#include<iostream>
#include<stdio.h>
#include<string.h>
#define N 110
//#define lowbit(x) x&(-x)
using namespace std;
int n,m;
int c[N][N][N];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int z,int val)
{
    while(x<=n)
    {
        int j=y;
        while(j<=n)
        {
            int k=z;
            while(k<=n)
            {
                c[x][j][k]+=val;
                k+=lowbit(k);
            }
            j+=lowbit(j);
        }
        x+=lowbit(x);
    }
}
int getsum(int x,int y,int z)
{
    int s=0;
    while(x>0)
    {
        int j=y;
        while(j>0)
        {
            int k=z;
            while(k>0)
            {
                s+=c[x][j][k];
                k-=lowbit(k);
            }
            j-=lowbit(j);
        }
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(c,0,sizeof c);
        int op,x1,y1,z1,x2,y2,z2;
        while(m--)
        {
            scanf("%d",&op);
            if(op)
            {
                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
                update(x2+1, y2+1, z2+1, 1);
                update(x1, y2+1, z2+1, 1);
                update(x2+1, y1, z2+1, 1);
                update(x2+1, y2+1, z1, 1);
                update(x1, y1, z2+1, 1);
                update(x2+1, y1, z1, 1);
                update(x1, y2+1, z1, 1);
                update(x1, y1, z1, 1);
            }
            else
            {
                scanf("%d%d%d",&x1,&y1,&z1);
                //cout<<"getsum(x1,y1,z1)="<<getsum(x1,y1,z1)<<endl;
                printf("%d\n",getsum(x1,y1,z1)&1);
            }
        }
    }
    return 0;
}