HDU 3584 cube-Three-dimensional tree Array

Give a three-dimensional array n * n, which is initially 0. Each operation has two operations:

1. Flip (x1, Y1, Z1)-> (X2, Y2, Z2)

0. query a [x] [y] [Z] (a is the array)

Solution: Number of maintenance operations in a tree array. an even number of operations is equivalent to no operation.

Update each update at eight locations:

. 8 binary numbers: 000,001,010,011,100,101,110,111 (I pushed it from two dimensions)

In fact, you don't need to use-1, because it will change parity.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>using namespace std;int c[105][105][105];int n;int lowbit(int x) { return x & (-x); }void modify(int x,int y,int z,int val){    for(int i=x;i<=n;i+=lowbit(i))        for(int j=y;j<=n;j+=lowbit(j))            for(int k=z;k<=n;k+=lowbit(k))                c[i][j][k] += val;}int getsum(int x,int y,int z){    int sum = 0;    for(int i=x;i>0;i-=lowbit(i))        for(int j=y;j>0;j-=lowbit(j))            for(int k=z;k>0;k-=lowbit(k))                sum += c[i][j][k];    return sum;}int main(){    int q,op;    int x,y,z,x1,y1,z1,x2,y2,z2;    while(scanf("%d%d",&n,&q)!=EOF)    {        memset(c,0,sizeof(c));        while(q--)        {            scanf("%d",&op);            if(op == 1)            {                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);                modify(x1,y1,z1,1);                modify(x1,y1,z2+1,-1);                modify(x1,y2+1,z1,-1);                modify(x1,y2+1,z2+1,1);                modify(x2+1,y1,z1,-1);                modify(x2+1,y1,z2+1,1);                modify(x2+1,y2+1,z1,1);                modify(x2+1,y2+1,z2+1,-1);            }            else            {                scanf("%d%d%d",&x,&y,&z);                int sum = getsum(x,y,z);                printf("%d\n",sum%2);            }        }    }    return 0;}

View code

 

HDU 3584 cube-Three-dimensional tree Array