Codeforces Round #188 (Div. 1) B. Ants 暴力

B. Ants

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/317/problem/B

Description

It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: every minute at each junction (x, y) containing at least four ants a group of four ants will be formed, and these four ants will scatter to the neighbouring junctions (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) — one ant in each direction. No other ant movements will happen. Ants never interfere with each other.

Scientists have put a colony of n ants into the junction (0, 0) and now they wish to know how many ants will there be at some given junctions, when the movement of the ants stops.

Input

First input line contains integers n (0 ≤ n ≤ 30000) and t (1 ≤ t ≤ 50000), where n is the number of ants in the colony and t is the number of queries. Each of the next t lines contains coordinates of a query junction: integers xi, yi ( - 109 ≤ xi, yi ≤ 109). Queries may coincide.

It is guaranteed that there will be a certain moment of time when no possible movements can happen (in other words, the process will eventually end).

Output

Print t integers, one per line — the number of ants at the corresponding junctions when the movement of the ants stops.

Sample Input

1 3
0 1
0 0
0 -1

Sample Output

0
1
0

HINT

题意

一个格子中的蚂蚁如果大于等于4个，这四个蚂蚁就会向四周扩散，扩散到(x+1,y)，(x-1,y)，(x,y+1)，(x,y-1)这四个格子

然后q次查询，每次查询(x,y)格子里面有多少个蚂蚁

题解：

蚂蚁最多为30000个，假设所有格子都是4只蚂蚁，那么也就最多30000/4=7500个格子

然后我们直接就好啦~

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 1000
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int a[maxn][maxn];
void solve(int x,int y)
{
    if(a[x][y]<)
        a[x][y]++;
    else
    {
        a[x][y]=;
        solve(x+,y);
        solve(x,y+);
        solve(x,y-);
        solve(x-,y);
    }
}
int main()
{
    int n=read(),q=read();
    for(int i=;i<n;i++)
        solve(,);
    for(int i=;i<q;i++)
    {
        int x=read(),y=read();
        if(x>||x<-||y>||y<-)
            cout<<""<<endl;
        else
            printf("%d\n",a[x+][y+]);
    }
}