ural1470 UFOs

UFOs

Time limit: 2.0 second
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 2 1 1 1 1 1 1 1 0 0 0 1 1 0 1 0 3 2 0 0 0 0 0 0 2 0 0 0 0 1 0 1 0 1 0 -2 2 0 0 0 1 1 1 3	0 1 4 2

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 ;

}