Codeforces Round #672 (Div. 2) A~D

只有当严格递减的时候才会达到冒泡排序最坏时间复杂度

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
	#endif
	LL _;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>_;
	
	while(_--){
		int n;
		cin>>n;
		vector<int> a(n);
		rep(i,0,n) cin>>a[i];
		vector<int> b = a;
		sort(all(a), greater<int>());
		bool flag = true;
		for(int i=0; i<n; i++){
			if(i&&a[i]==a[i-1])
				flag = false;
			if(a[i]!=b[i])
				flag  =false;
		}
		if(flag) 
			cout<<"NO"<<endl;
		else
			cout<<"YES"<<endl;
	}
	return 0;
}

按照最高位进行分组然后统计答案就行了

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
	#endif
	LL _;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>_;
	
	while(_--){
		LL n;
		cin>>n;
		LL a;
		map<LL, LL> cnt;
		rep(i,0,n){
			cin>>a;
			LL k=0;
			
			while(a>=(1LL<<k)){
				k++;
			}
			cnt[k]++;
		}
		LL  ans = 0;
		for(auto e:cnt){
			ans += (e.se*(e.se-1))/2;
		}
		cout<<ans<<endl;
	}
	return 0;
}

直接dp，记录前面为奇数长度和为偶数长的的最大值就可以了

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()


int main(int argc, char const *argv[])
{
//	#define DEBUG
    	#ifdef DEBUG
		freopen("1.dat","r",stdin);
	#endif
	LL _;
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>_;
	
	while(_--){
		int n,q;
		cin>>n>>q;
		LL ans = 0LL;
		vector<LL> a(n);
		rep(i,0,n) {
			cin>>a[i];
		}
		set<LL> odd, even;
		odd.insert(a[0]);
		for(int i=1; i<n; i++){
			odd.insert(a[i]);
			if(even.size()){
				odd.insert(*(--even.end())+a[i]);
			}
			even.insert(*(--odd.end())-a[i]);
		}
		if(even.size())
			ans = max(ans,*(--even.end()));
		ans = max(ans,*(--odd.end()));
		cout<<ans<<endl;
	}
	return 0;
}

大佬给的思路，如果a[i]-a[i+1]大于零，那么这一组就要选
于是动态的更新就好了

#include<bits/stdc++.h>
#define LL long long
using namespace std;

int main(int argc, char const *argv[])
{
	int _;
	cin>>_;
	while(_--){
		
		int n,q;
		cin>>n>>q;
		vector<int> a(n+10,0);
		LL ans = 0;
		for(int i=1; i<=n; ++i){
			cin>>a[i];
		}
		for(int i=0; i<=n; i++){
			ans += max(0, a[i]-a[i+1]);
		}
		cout<<ans<<endl;
		int x,y;
		while(q--){
			cin>>x>>y;

			if(x==y){
				cout<<ans<<endl;
				continue;
			}
			ans -= max(a[x-1]-a[x], 0);
			ans -= max(a[x]-a[x+1], 0);
			ans -= max(a[y]-a[y+1], 0);
			if(x+1!=y)
				ans -= max(a[y-1]-a[y],0);

			swap(a[x], a[y]);
			ans += max(a[x-1]-a[x], 0);
			ans += max(a[x]-a[x+1], 0);
			ans += max(a[y]-a[y+1], 0);
			if(x+1!=y)
				ans += max(a[y-1]-a[y],0);
			cout<<ans<<endl;
		}
	}
	return 0;
}

按照l从小到大排序之后
那么用优先队列维护r的值
对于每一个新的区间，抛弃掉优先队列里r小于a[i]的l的那些区间就可以
剩下就是求组合数了(学到了Lucas定理和小费马定理)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()

const int MAXN = 3e5+10;

LL mod = 998244353;

LL pre[MAXN];

LL qpow(LL x, LL n){
	LL res = 1LL;
	while(n){
		if(n&1) res = (res*x)%mod;
		n>>=1;
		x = (x*x)%mod;
	}
	return res%mod;
}

LL C(LL n, LL k){
	if(k==0) return 1;
	return (pre[n]%mod*qpow((pre[k]*pre[n-k])%mod, mod-2)%mod)%mod;
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int n,k;
	cin>>n>>k;
	pre[0] = 1;
	rep(i,1,MAXN){
		pre[i] = (pre[i-1]*i)%mod;
	}
	vector<pair<int, int> > a(n);
	rep(i, 0, n) cin>>a[i].fi>>a[i].se;
	sort(all(a));
	LL ans = 0LL;
	priority_queue<int, vector<int>, greater<int> > pq;
	for(int i=0; i<n; i++){
		while(pq.size()&&a[i].fi>pq.top()){
			pq.pop();
		}
		if(pq.size()>=k-1){
			ans = (ans+C(pq.size(), k-1));
		}
		pq.push(a[i].se);
	}
	cout<<ans%mod<<endl;
	return 0;
}

我好菜啊