Cube
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
1 /*AC代码*/ 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 6 using namespace std; 7 8 const int N=110; 9 10 int n,m,arr[N][N][N]; 11 12 int lowbit(int x) 13 { 14 return x&(-x); 15 } 16 17 void update(int i,int j,int k,int val) 18 { 19 while(i<=n) 20 { 21 int tmpj=j; 22 while(tmpj<=n) 23 { 24 int tmpk=k; 25 while(tmpk<=n) 26 { 27 arr[i][tmpj][tmpk]+=val; 28 tmpk+=lowbit(tmpk); 29 } 30 tmpj+=lowbit(tmpj); 31 } 32 i+=lowbit(i); 33 } 34 } 35 36 int Sum(int i,int j,int k) 37 { 38 int ans=0; 39 while(i>0) 40 { 41 int tmpj=j; 42 while(tmpj>0) 43 { 44 int tmpk=k; 45 while(tmpk>0) 46 { 47 ans+=arr[i][tmpj][tmpk]; 48 tmpk-=lowbit(tmpk); 49 } 50 tmpj-=lowbit(tmpj); 51 } 52 i-=lowbit(i); 53 } 54 return ans; 55 } 56 57 int main() 58 { 59 60 //freopen("input.txt","r",stdin); 61 62 while(~scanf("%d%d",&n,&m)) 63 { 64 memset(arr,0,sizeof(arr)); 65 int x1,y1,z1,x2,y2,z2; 66 int op; 67 while(m--) 68 { 69 scanf("%d",&op); 70 if(op==1) 71 { 72 scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2); 73 update(x2+1, y2+1, z2+1, 1); 74 update(x1, y2+1, z2+1, 1); 75 update(x2+1, y1, z2+1, 1); 76 update(x2+1, y2+1, z1, 1); 77 update(x1, y1, z2+1, 1); 78 update(x2+1, y1, z1, 1); 79 update(x1, y2+1, z1, 1); 80 update(x1, y1, z1, 1); 81 } 82 else 83 { 84 scanf("%d%d%d",&x1,&y1,&z1); 85 printf("%d ",Sum(x1,y1,z1)&1); 86 //该点的值就是sum(x,y) 87 } 88 } 89 } 90 return 0; 91 }
感觉三维应该和二维差不多 变一下下就好了
但是我就是wa了 囧 照例，下面的代码 请无视
1 /*wa代码*/ 2 #include<iostream> 3 4 using namespace std; 5 6 int N; 7 int c[105][105][105]; 8 9 int lowbit( int x ) 10 { 11 return x & (-x); 12 } 13 14 void update( int x, int y, int z, int delta ) 15 { 16 int i, j , k; 17 for(i=x; i<=N; i+=lowbit(i)) 18 { 19 for(j=y; j<=N; j+=lowbit(j)) 20 { 21 for(k=z ; k<=N; k+=lowbit(j)) 22 c[i][j][k] += delta; 23 } 24 } 25 } 26 27 int sum( int x, int y , int z ) 28 { 29 int res = 0, i, j , k; 30 for(i=x; i>0; i-=lowbit(i)) 31 { 32 for(j=y; j>0; j-=lowbit(j)) 33 { 34 for(k=z; k>0; k-=lowbit(j)) 35 { 36 res += c[i][j][k]; 37 } 38 } 39 } 40 return res; 41 } 42 43 44 void init () 45 { 46 int i,j,k; 47 for(i=0;i<=N;i++) 48 for(j=0;j<=N;j++) 49 for(k=0;k<=N;k++) 50 c[i][j][k]=0; 51 } 52 53 int main() 54 { 55 int x1,y1,z1,x2,y2,z2; 56 int temp,p; 57 58 while(scanf("%d%d%",&N,&p)!=EOF) 59 { 60 init (); 61 while(p--) 62 { 63 scanf("%d",&temp); 64 if(temp==1) 65 { 66 // printf("input x1~z2 "); 67 68 scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2); 69 update(x2+1, y2+1, z2+1, 1); 70 update(x1, y2+1, z2+1, 1); 71 update(x2+1, y1, z2+1, 1); 72 update(x2+1, y2+1, z1, 1); 73 update(x1, y1, z2+1, 1); 74 update(x2+1, y1, z1, 1); 75 update(x1, y2+1, z1, 1); 76 update(x1, y1, z1, 1); 77 78 } 79 else if(temp==0) 80 { 81 scanf("%d%d%d",&x1,&y1,&z1); 82 printf("%d ",sum(x1,y1,z1)&1); 83 } 84 } 85 } 86 87 return 0 ; 88 }