UFOs
Time limit: 2.0 second
Memory limit: 64 MB
Memory limit: 64 MB
Vasya is a ufologist and his duties include observing Unidentified Flying Objects (UFOs) in the part of space bounded by a cube N × N × N. The cube is divided into cubic sectors 1 × 1 × 1. During the observation, the following events may happen:
- several new UFOs emerge in a certain sector;
- several UFOs disappear in a certain sector;
- Vasya's boss may ask him how many UFOs there are in a part of space consisting of several sectors.
At the moment when Vasya starts his observations there are no UFOs in the whole space.
Input
The first line contains an integer N (1 ≤ N ≤ 128). The coordinates of sectors are integers from 0 to N–1.
Then there are entries describing events, one entry per line. Each entry starts with a number M.
- If M is 1, then this number is followed by four integers x (0 ≤ x < N), y (0 ≤ y < N), z (0 ≤ z < N), K (–20000 ≤ K ≤ 20000), which are coordinates of a sector and the change in the number of UFOs in this sector. The number of UFOs in a sector cannot become negative.
- If M is 2, then this number is followed by six integers x1, y1, z1, x2, y2, z2 (0 ≤ x1 ≤ x2 <N, 0 ≤ y1 ≤ y2 < N, 0 ≤ z1 ≤ z2 < N), which mean that Vasya must compute the total number of UFOs in sectors (x, y, z) belonging to the volume: x1 ≤ x ≤ x2, y1 ≤ y ≤ y2, z1 ≤ z ≤z2.
- If M is 3, it means that Vasya is tired and goes to sleep. This entry is always the last one.
The number of entries does not exceed 100002.
Output
For each query, output in a separate line the required number of UFOs.
Sample
| input | output |
|---|---|
2 |
0 |
分析：三维树状数组，求和时类似于容斥；
代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
int n,m,k,t;
ll a[][][];
void add(int x,int y,int z,int w)
{
for(int i=x;i<=n;i+=(i&(-i)))
for(int j=y;j<=n;j+=(j&(-j)))
for(int k=z;k<=n;k+=(k&(-k)))
a[i][j][k]+=w;
}
ll get(int x,int y,int z)
{
ll ans=;
for(int i=x;i;i-=(i&(-i)))
for(int j=y;j;j-=(j&(-j)))
for(int k=z;k;k-=(k&(-k)))
ans+=a[i][j][k];
return ans;
}
int main()
{
int i,j;
scanf("%d",&n);
while(~scanf("%d",&m)&&m!=)
{
if(m==)
{
int b[];
rep(i,,)scanf("%d",&b[i]);
add(++b[],++b[],++b[],b[]);
}
else
{
int b[];
rep(i,,)scanf("%d",&b[i]),b[i]++;
printf("%lld\n",get(b[],b[],b[])
-get(b[]-,b[],b[])-get(b[],b[]-,b[])-get(b[],b[],b[]-)
+get(b[]-,b[]-,b[])+get(b[],b[]-,b[]-)+get(b[]-,b[],b[]-)
-get(b[]-,b[]-,b[]-));
}
}
//system("Pause");
return ;
}