2018 ICPC 沈阳 G.Best ACMer Solves the Hardest Problem（预处理 思维）

题目链接：https://nanti.jisuanke.com/t/A2168

分析

这题关键在怎么将 k 拆分成两个可以开方的数，可以先预处理出可能的对数，然后直接调用，同时一定要注意越界的问题。

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll mod = 6000;
int n,m,t;
map<pair<int,int>,int> mp;
ll ans;
vector<pair<int,int> > g[10000007];

bool check(int x,int y)
{
	if(x <= 6000 && x > 0 && y > 0 && y <= 6000) return 1;
	return 0;
}

void init()
{
	for(int i=0;i<=6000;i++)
	{
		for(int j=0;j<=6000;j++)
		{
			if(i * i + j * j > 1e7) break;
			g[i * i + j * j].push_back({i, j});
		}
	}
}

int main()
{
	init();
	scanf("%d",&t);
	for(int T=1;T<=t;T++)
	{
		mp.clear();
		ans = 0;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
		{
			int x,y,w;
			scanf("%d%d%d",&x,&y,&w);
			mp[{x, y}] = w;
		}
		printf("Case #%d:\n",T);
		while(m--)
		{
			int op,x,y,w,k;
			scanf("%d%d%d",&op,&x,&y);
			x = (x + ans) % mod + 1;
			y = (y + ans) % mod + 1;
			if(op == 1)
			{
				scanf("%d",&w);
				mp[{x, y}] = w;
			}
			if(op == 2)
			{
				mp[{x, y}] = 0;
			}
			if(op == 3)
			{
				scanf("%d%d",&k,&w);
				if(k == 0)
				{
					mp[{x, y}] += w;
					continue;
				}
				for(int i=0;i<g[k].size();i++)
				{
					int t1 = g[k][i].first;
					int t2 = g[k][i].second;
					if(t1 == 0)
					{
						if(check(x - t1, y + t2) && mp[{x - t1, y + t2}] > 0) mp[{x - t1, y + t2}] += w;
						if(check(x - t1, y - t2) && mp[{x - t1, y - t2}] > 0) mp[{x - t1, y - t2}] += w;
					}
					else if(t2 == 0)
					{
						if(check(x + t1, y + t2) && mp[{x + t1, y + t2}] > 0) mp[{x + t1, y + t2}] += w;
						if(check(x - t1, y + t2) && mp[{x - t1, y + t2}] > 0) mp[{x - t1, y + t2}] += w;
					}
					else
					{
						if(check(x + t1, y + t2) && mp[{x + t1, y + t2}] > 0) mp[{x + t1, y + t2}] += w;
						if(check(x - t1, y + t2) && mp[{x - t1, y + t2}] > 0) mp[{x - t1, y + t2}] += w;
						if(check(x + t1, y - t2) && mp[{x + t1, y - t2}] > 0) mp[{x + t1, y - t2}] += w;
						if(check(x - t1, y - t2) && mp[{x - t1, y - t2}] > 0) mp[{x - t1, y - t2}] += w;
					}
				}
			}
			if(op == 4)
			{
				ans = 0;
				scanf("%d",&k);
				if(k == 0)
				{
					printf("%lld\n",mp[{x, y}]);
					ans = mp[{x, y}];
					continue;
				}
				for(int i=0;i<g[k].size();i++)
				{
					int t1 = g[k][i].first;
					int t2 = g[k][i].second;
					if(t1 == 0)
					{
						if(check(x - t1, y + t2) && mp[{x - t1, y + t2}] > 0) ans += mp[{x - t1, y + t2}];
						if(check(x - t1, y - t2) && mp[{x - t1, y - t2}] > 0) ans += mp[{x - t1, y - t2}];
					}
					else if(t2 == 0)
					{
						if(check(x + t1, y + t2) && mp[{x + t1, y + t2}] > 0) ans += mp[{x + t1, y + t2}];
						if(check(x - t1, y + t2) && mp[{x - t1, y + t2}] > 0) ans += mp[{x - t1, y + t2}];
					}
					else
					{
						if(check(x - t1, y + t2) && mp[{x - t1, y + t2}] > 0) ans += mp[{x - t1, y + t2}];
						if(check(x - t1, y - t2) && mp[{x - t1, y - t2}] > 0) ans += mp[{x - t1, y - t2}];
						if(check(x + t1, y - t2) && mp[{x + t1, y - t2}] > 0) ans += mp[{x + t1, y - t2}];
						if(check(x + t1, y + t2) && mp[{x + t1, y + t2}] > 0) ans += mp[{x + t1, y + t2}];
					}
				}
				printf("%lld\n",ans);
			}
		}
	}
	return 0;
}