【A-D】Codeforces Round #812 (Div. 2)参考代码

A. Traveling Salesman Problem

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#define ll long long
using namespace std;
const int mod=998244353;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}

void solve()
{
	int n=read();
	int x=0,xx=0,y=0,yy=0;
	for(int i=1;i<=n;i++)
	{
		int a=read(),b=read();
		if(a==0)
		{
			if(b>0)y=max(y,b);
			else yy=min(yy,b);
		}
		else
		{
			if(a>0)x=max(x,a);
			else xx=min(xx,a);
		}
	} 
	printf("%lld\n",(ll)2*(y+x-xx-yy));
}

int main()
{
	int T=read();
	while(T--)solve(); 
	return 0;
}


B. Optimal Reduction

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#define ll long long
using namespace std;
const int mod=998244353;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}
const int N=1e5+10;
int a[N]; 
void solve()
{
	int n=read();
	for(int i=1;i<=n;i++)a[i]=read();
	bool ans1=0,ans2=0;
	int id;
	for(int i=1;i<n;i++)
	{
		if(a[i]>a[i+1])
		{
			if(ans1==0)id=i;
			ans1=1;
		}
	}
		
	if(ans1==0)
	{
		printf("YES\n");
		return;
	}
	for(int i=id;i<n;i++)
		if(a[i]<a[i+1])ans2=1;
	
	if(ans2==0)
	{
		printf("YES\n");
		return;		
	}
	printf("NO\n");
}

int main()
{
	int T=read();
	while(T--)solve();
	return 0;
}



C. Build Permutation

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#define ll long long
using namespace std;
const int mod=998244353;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}
const int N=1e5+10;
ll fuck[510];
ll ans[N];
bool v[N];
void solve()
{
	memset(v,0,sizeof(v));
	int n=read();
	int id;
	for(int i=1;i<500;i++)
	{
		ll x=2*n;
		if(fuck[i]<x&&fuck[i+1]>x)id=i;
	}
	
	for(int i=n;i>=1;i--)
	{
		while(1)
		{
			ll x=fuck[id]-i+1;
			if(x<0||x>=n||v[x])
			{
				id--;
				continue;
			}
			ans[i]=x;
			v[x]=1;
			break;
		}
	}
	for(int i=1;i<=n;i++)printf("%lld ",ans[i]);
	printf("\n"); 
}

int main()
{
	for(int i=1;i<=500;i++)fuck[i]=i*i;
	int T=read();
	while(T--)solve();
	return 0;
}



D. Tournament Countdown

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#define ll long long
using namespace std;
const int mod=998244353;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}
const int N=2e5+10;
int fuck[20];
int a[N];
void ask(int x,int y){cout<<"? "<<x<<" "<<y<<endl;}
void ans(int x){cout<<"! "<<x<<endl;}
int qqq(int a,int b,int c,int d)
{
	ask(a,c);
	int p=read();
	if(p==0)
	{
		ask(b,d);
		int q=read();
		if(q==1)return b;
		else return d; 
	}
	if(p==1)
	{
		ask(a,d);
		int q=read();
		if(q==1)return a;
		else return d;
	}
	if(p==2)
	{
		ask(b,c);
		int q=read();
		if(q==1)return b;
		else return c;
	}
}
void solve()
{
	vector<int>v;
	int n=read();
	for(int i=1;i<=fuck[n];i++)v.push_back(i);
	while(n>1)
	{
		int g=fuck[n]/4;
		for(int i=1;i<=fuck[n];i++)a[i]=v[i-1];
		v.clear();
		for(int i=1;i<=g;i++)
		{
			int p=qqq(a[(i-1)*4+1],a[(i-1)*4+2],a[(i-1)*4+3],a[(i-1)*4+4]);
			v.push_back(p);
		}
		n-=2;
	}
	if(n==1)
	{	
		ask(v[0],v[1]);
		int q=read();
		if(q==1)ans(v[0]);
		else ans(v[1]);
	}
	else ans(v[0]);
}

int main()
{
	fuck[1]=2;
	for(int i=2;i<=19;i++)fuck[i]=fuck[i-1]*2;
	int T=read();
	while(T--)solve();
	return 0;
}



评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值