第一题：纯模拟题

#include<bits/stdc++.h>
using namespace std;
int n = 5;
int a[6],b[6],aa[6],bb[6];
int main()
{
	for(int i = 1;i<=5;i++)cin>>a[i];
	for(int i = 1;i<=5;i++)cin>>aa[i];
	for(int i = 1;i<=5;i++)cin>>b[i];
	for(int i = 1;i<=5;i++)cin>>bb[i];
	bool flag = true;
	int cnt1 = 1, cnt2 = 1; 
	while(cnt1 <=5 && cnt2 <=5)
	{
		//轮到小易 
		if(flag)
		{
			bb[cnt2]-=a[cnt1];
			flag = false;
			if(bb[cnt2]<=0)cnt2++;
		}
		else
		{
			aa[cnt1]-=b[cnt2];
			flag = true;
			if(aa[cnt1]<=0)cnt1++;
		}	
	}
	if(cnt1 <=5)
	{
		cout << "win"<<endl;
		cout << n - cnt1 + 1 <<endl;
	}
	else
	{
		cout << "lose"<<endl;
		cout << n - cnt2 + 1 <<endl;
	}
}

第二题：递归求组合数 然后再进行计算最大值就好了

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 410; 
int n;
struct jianzhu
{
	int val;
	int a;
	int b;
	int c;
}p[N];
int sum = 0;
void dfs(int val, int aa,int bb,int cc,int start)
{	
	if(start>n)
	{
		return;
	}
	for(int i = start; i <=n;i++)
	{
		if(aa>=p[i].a && bb>=p[i].b && cc>=p[i].c)
		{
			sum = max(sum,val + p[i].val);
			dfs(val + p[i].val,  aa - p[i].a,bb-p[i].b,cc-p[i].c,i + 1);
		}
	}
}
signed main()
{
	int A,B,C;
	cin>>n>>A>>B>>C;
	for(int i = 1;i<=n;i++)
	{
		int a,b,c,v;
		cin>>a>>b>>c>>v;
		p[i]={v,a,b,c};
	}
	dfs(0,A,B,C,1);
	cout << sum <<endl;
	return 0;
}

第三题：先把坐标整体偏移2000，然后做一次二维差分数组就好了，一开始以为是哈希，推了半天没推了，后面发现，这不裸着的二维差分数组嘛！

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int n,q;
int b[5000][5000];
struct Dian
{
	int x,y,r;
}dian[N];
void add(int x1,int y1,int x2,int y2,int c)
{
	b[x1][y1]+=c;
	b[x1][y2+1]-=c;
	b[x2+1][y1]-=c;
	b[x2+1][y2+1]+=c;
}
signed main()
{
	cin>>n;
	for(int i =1;i<=n;i++)
	{
		int x,y,r;
		cin>>x>>y>>r;
		x+=2000,y+=2000;
		dian[i] = {x,y,r};
		add(x-r,y-r,x+r,y+r,1);
		
	}
	for(int i = 1;i<=5000;i++)
	{
		for(int j = 1;j<=5000;j++)
		{
			b[i][j] = b[i][j] + b[i-1][j] + b[i][j-1] - b[i-1][j-1];
		}
	}
	cin>>q;
	while(q--)
	{
		int x0,y0;
		cin>>x0>>y0;
		x0+=2000,y0+=2000;
		cout <<b[x0][y0]<<endl;
	}
	return 0;
}

今天的题目好简单阿，一小时ak，好久没那么爽了，感谢网易互娱~~