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.

[b]Input[/b]

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).
[b]Output[/b]

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

[b]Sample Input[/b]

1 3
0 1
0 0
0 -1

[b]Sample Output[/b]

0
1
0

HINT

[b]题意[/b]

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

[b]题解：[/b]

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~

[b]代码：[/b]

#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=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//**************************************************************************************

int a[maxn][maxn];
void solve(int x,int y)
{
if(a[x][y]<3)
a[x][y]++;
else
{
a[x][y]=0;
solve(x+1,y);
solve(x,y+1);
solve(x,y-1);
solve(x-1,y);
}
}
int main()
{
int n=read(),q=read();
for(int i=0;i<n;i++)
solve(500,500);
for(int i=0;i<q;i++)
{
int x=read(),y=read();
if(x>100||x<-100||y>100||y<-100)
cout<<"0"<<endl;
else
printf("%d\n",a[x+500][y+500]);
}
}