牛客小白月赛63 ABCD

https://ac.nowcoder.com/acm/contest/49030

这套题目质量挺好的，E过了200＋，状态不佳，改天补补

A-子序列的权值最小值

输入 
6
1 1 4 5 1 4
输出 
0

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=2e7+10,M=2023;
const LL mod=1e9+7;
const double PI=3.1415926535;
#define endl '\n'
LL a[N],b[N];  
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
    	LL n;
    	cin>>n;
    	LL sum=0;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>a[i];
    		if(i==1) sum=a[i];
    		else sum&=a[i];
		}
		cout<<sum<<endl;
    }
    return 0;
}

B-魔导师晨拥

输入 
5 6
1 2 3 4 5
输出 
36

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=2e7+10,M=2023;
const LL mod=1e9+7;
const double PI=3.1415926535;
#define endl '\n'
LL a[N],b[N];  
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
    	LL n,m;
    	cin>>n>>m;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>a[i];
		}
		LL beg=2;
		LL ans=0;
    	       for(int i=1;i<=m;i++)
    	       {
    		for(int j=1;j<=n;j++)
    		{
    			a[j]-=beg;
    			if(a[j]==0) beg++;
			}
			ans+=beg;
		}
		cout<<ans<<endl;
    }
    return 0;
}

C-GCPC总决赛

输入 
2
0 2
1 3
输出 
0 1 1

数据范围小，直接暴力即可

方法一

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=1e6+10,M=5005;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
LL a[N],b[N];
LL num[N];
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
    	LL n;
    	cin>>n;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>a[i];
    		num[i]=i;
		}
		for(int i=1;i<=n;i++)
		{
			cin>>b[i];
		}
		LL sum1=0,sum2=0,sum3=0;
		do
		{
			LL x=0,y=0,z=0;
			for(int i=1;i<=n;i++)
			{
				int aa=a[num[i]],bb=b[i];
				if(aa>bb) x++;
				else if(aa<bb) y++;
				else z++;
			}
			if(x>y) sum1++;
			else if(x<y) sum2++;
			else sum3++;
		}
		while(next_permutation(num+1,num+1+n));
		cout<<sum1<<" "<<sum2<<" "<<sum3<<endl;
	}
    return 0;
}

方法二

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=1e6+10,M=5005;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
LL n,sum1=0,sum2=0,sum3=0;
LL a[N],b[N];
bool st[N];
vector<LL> v;
void dfs(int idx)
{
	if(idx>n)
	{
		LL x=0,y=0,z=0;
		for(int i=0;i<v.size();i++)
		{
			if(v[i]>b[i+1]) x++;
			else if(v[i]<b[i+1]) y++;
			else z++;
		}
		if(x>y) sum1++;
		else if(x<y) sum2++;
		else sum3++;
		return ;
	}
	for(int i=1;i<=n;i++)
	{
		if(st[i]==0)
		{
			st[i]=1;
			v.push_back(a[i]);
			dfs(idx+1);
			v.pop_back();
			st[i]=0;
		}
	}
}
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
    	cin>>n;
    	for(int i=1;i<=n;i++)
    	{
    		cin>>a[i];
		}
		for(int i=1;i<=n;i++)
		{
			cin>>b[i];
		}
		dfs(1);
		cout<<sum1<<" "<<sum2<<" "<<sum3<<endl;
	}
    return 0;
}

D-Ginger的大花环

输入 
5 4
1 2 3 4
输出 
7 

输入 
3 1
1
输出 
Ginger666

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18,MINN=-MAXN,INF=0x3f3f3f3f;
const LL N=1e6+10,M=5005;
const LL mod=100000007;
const double PI=3.1415926535;
#define endl '\n'
LL a[N];
int main()
{
    cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
    LL T=1;
    //cin>>T;
    while(T--)
    {
    	LL n,k;
    	cin>>n>>k;
    	for(int i=1;i<=k;i++)
    	{
    		cin>>a[i];
		}
		if(k<=1) cout<<"Ginger666"<<endl;
		else
		{
			sort(a+1,a+1+k);
			if(n%3==0) cout<<(n/3)*(2*a[1]+a[2]);
			else if(n%3==1) cout<<(n/3)*(2*a[1]+a[2])+a[2];
			else if(n%3==2) cout<<(n/3)*(2*a[1]+a[2])+a[1]+a[2];
		}
	}
    return 0;
}