Mynavi Programming Contest 2021（AtCoder Beginner Contest 201）ABCDE_do you know the second highest mountain?c++

Mynavi Programming Contest 2021（AtCoder Beginner Contest 201）

A - Tiny Arithmetic Sequence

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define inf 0x7fffffff
#define ll long long
//#define llu unsigned long long
//#define int long long
//#define int __int128
//#define double long double
//#define double long long
#define re register int
//#define void inline void
#define eps 1e-8
//#define mod 1e9+7
#define ls(p) p<<1
#define rs(p) p<<1|1
#define pi acos(-1.0)
#define pb push_back
#define mk make_pair
#define P pair < int , int >
using namespace std;
const int mod=1e9+7;
const int M=1e8;
const int N=5e5+5;//??????.???? 4e8
int a[N]; 
void solve()
{
	cin>>a[1]>>a[2]>>a[3];
	sort(a+1,a+3+1);
	if(a[1]+a[3]==2*a[2])  puts("Yes");
	else  puts("No"); 
}
signed main()
{
    int T=1;
//    cin>>T;
    for(int index=1;index<=T;index++)
    {
        solve();
//        puts("");
    }
    return 0;
}
/*
 
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*/
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B - Do you know the second highest mountain?

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define inf 0x7fffffff
#define ll long long
//#define llu unsigned long long
//#define int long long
//#define int __int128
//#define double long double
//#define double long long
#define re register int
//#define void inline void
#define eps 1e-8
//#define mod 1e9+7
#define ls(p) p<<1
#define rs(p) p<<1|1
#define pi acos(-1.0)
#define pb push_back
#define mk make_pair
#define P pair < int , int >
using namespace std;
const int mod=1e9+7;
const int M=1e8;
const int N=5e5+5;//??????.???? 4e8
struct node
{
	int x;
	string s;
}e[N];
int n;
bool cmp(node i,node j)
{
	if(i.x!=j.x)  return i.x>j.x;
	return i.s<j.s;
}
void solve()
{
	cin>>n;
	for(re i=1;i<=n;i++)  cin>>e[i].s>>e[i].x;
	sort(e+1,e+n+1,cmp);
	cout<<e[2].s;
}
signed main()
{
    int T=1;
//    cin>>T;
    for(int index=1;index<=T;index++)
    {
        solve();
//        puts("");
    }
    return 0;
}
/*
 
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*/
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C - Secret Number

这大概是最恶心的一道题了，公式推了无数次，WA了又WA

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define inf 0x7fffffff
#define ll long long
//#define llu unsigned long long
//#define int long long
//#define int __int128
//#define double long double
//#define double long long
#define re register int
//#define void inline void
#define eps 1e-8
//#define mod 1e9+7
#define ls(p) p<<1
#define rs(p) p<<1|1
#define pi acos(-1.0)
#define pb push_back
#define mk make_pair
#define P pair < int , int >
using namespace std;
const int mod=1e9+7;
const int M=1e8;
const int N=5e5+5;//??????.???? 4e8
char s[N];
int n,a1,a2,a3;
int ans=1,cnt=1,ret;
void solve()
{
	for(re i=1;i<=4;i++)   cnt*=i;
	scanf("%s",s+1);
	n=strlen(s+1);
	for(re i=1;i<=n;i++)  if(s[i]=='o')  a1++;
	else if(s[i]=='?')  a3++;
	if(a1>4)
	{
		puts("0");
		return;
	} 
	if(a1==4)
	{
		cout<<cnt<<endl;
		return;
	}
	if(a1==0)
	{
		for(re i=1;i<=4;i++)  ans*=a3;
		cout<<ans<<endl;
		return;
	}
	if(a1==1)
	{
		cout<<4*a3*a3*a3+6*a3*a3+4*a3+1<<endl;
		return;
	}
	if(a1==2)
	{
//		cout<<8+24*a3*a3+2*12*a3<<endl;
		cout<<14+12*a3*a3+24*a3<<endl;
		return;
	}
	if(a1==3)
	{
		ret+=36;
		ret+=24*a3;
		cout<<ret<<endl;
		return;
	}
}
signed main()
{
    int T=1;
//    cin>>T;
    for(int index=1;index<=T;index++)
    {
        solve();
//        puts("");
    }
    return 0;
}
/*
 
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*/
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D - Game in Momotetsu World

好吧，最开始想了很久都没想到要反向推。
问：为什么取负，其实无所谓，主要是反向推取负便与思考。

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define inf 0x7fffffff
#define ll long long
//#define llu unsigned long long
#define int long long
//#define int __int128
//#define double long double
//#define double long long
#define re register int
//#define void inline void
#define eps 1e-8
//#define mod 1e9+7
#define ls(p) p<<1
#define rs(p) p<<1|1
#define pi acos(-1.0)
#define pb push_back
#define mk make_pair
#define P pair < int , int >
using namespace std;
const int mod=1e9+7;
const int M=1e8;
const int N=2e3+5;//??????.???? 4e8
int f[N][N];
char s[N][N];
int n,m;
void solve()
{
	cin>>n>>m;
	for(re i=1;i<=n;i++)  scanf("%s",s[i]+1);
	for(re i=n;i>=1;i--)  for(re j=m;j>=1;j--)
	{
		if(i==n&&j==m)  continue;
		int a1=-1e18,a2=-1e18;
		if(i!=n)
		{
			a1=-f[i+1][j];
			if(s[i+1][j]=='+')  a1++;
			else  a1--;
		}
		if(j!=m)
		{
			a2=-f[i][j+1];
			if(s[i][j+1]=='+')  a2++;
			else  a2--;
		}
		f[i][j]=max(a1,a2); 
	}
	if(f[1][1]>0)  puts("Takahashi");
	else if(f[1][1]==0)  puts("Draw");
	else  puts("Aoki");
}
signed main()
{
    int T=1;
//    cin>>T;
    for(int index=1;index<=T;index++)
    {
        solve();
//        puts("");
    }
    return 0;
}
/*
 
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1
 
 
 
 
*/
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E - Xor Distances

最先没想到怎么做，后来听朋友说，计算按位每一位的贡献。。。。

#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>
#define inf 0x7fffffff
#define ll long long
//#define llu unsigned long long
#define int long long
//#define int __int128
//#define double long double
//#define double long long
#define re register int
//#define void inline void
#define eps 1e-8
//#define mod 1e9+7
#define ls(p) p<<1
#define rs(p) p<<1|1
#define pi acos(-1.0)
#define pb push_back
#define mk make_pair
#define P pair < int , int >
using namespace std;
const int mod=1e9+7;
const int M=1e8;
const int N=5e5+5;//??????.???? 4e8
int n;
struct node
{
	int ver,edge,next;
}e[N];
int tot,head[N];
int d[N],f[N];
int ret=1,ans;
void add(int x,int y,int z)
{
	e[++tot].ver=y;
	e[tot].edge=z;
	e[tot].next=head[x];
	head[x]=tot;
}
void addedge(int x,int y,int z)
{
	add(x,y,z);add(y,x,z);
}
void dfs(int x,int pre)
{
	for(re i=head[x];i;i=e[i].next)
	{
		int y=e[i].ver;
		int z=e[i].edge;
		if(y==pre)  continue;
		d[y]=d[x]^z;
		dfs(y,x);
	}
}
void solve()
{
	cin>>n;
	for(re i=1;i<n;i++)
	{
		int x,y,z;
		scanf("%lld%lld%lld",&x,&y,&z);
		addedge(x,y,z);
	}
	dfs(1,1);
	for(re i=0;i<=60;i++)
	{
		int cnt=0;
		for(re j=1;j<=n;j++)  if((d[j]>>i)&1)  cnt++;
		ans=(ans+(ret*cnt)%mod*(n-cnt))%mod;
		ret=(ret*2)%mod;
	}
	cout<<ans<<endl;
}
signed main()
{
    int T=1;
//    cin>>T;
    for(int index=1;index<=T;index++)
    {
        solve();
//        puts("");
    }
    return 0;
}
/*

5 10
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1



*/
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