Codeforces Round #723 (Div. 2) 部分题解

一发过了 D ，nice

A

思维题，注意到排序后，将 \(a_{n+i}\) 插入到 \(a_i\) 前即可。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

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

int a[30];
int main(){
	int T; cin>>T;
	while(T--){
		int n; cin>>n;
		rep(i,1,2*n) cin>>a[i];
		
		sort(a+1, a+1+2*n);
		
		rep(i,1,n) cout<<a[i]<<' '<<a[i+n]<<' ';
		cout<<endl;
	}
    return 0;
}

B

分析：数学题，推推柿子即可

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

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

int main(){
	int T; cin>>T;
	while(T--){
		int x; cin>>x;
		int r=x%11;
		if(!r) puts("YES");
		else{
			x-=r*111;
			if(x<0) puts("NO");
			else if(x==0) puts("YES");
			else{
				if(x%11==0) puts("YES");
				else puts("NO");
			}
		}
	}
    return 0;
}

C

反悔贪心
细节见代码：

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

int main(){
	int n; cin>>n;
	priority_queue<ll, vector<ll>, greater<ll>> pq;
	
	ll cur=0;
	rep(i,1,n){
		ll x; cin>>x;
		pq.push(x), cur+=x;
		
		while(cur<0 && pq.size()){
			cur-=pq.top();
			pq.pop();
		}
	}
	cout<<pq.size()<<endl;
	
    return 0;
}

D

发现相同的字母总是连在一起能得到最优解，所以暴力 \(4!\) 枚举排列，然后统计逆序对即可。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=2e5+5;
int a[5], cnt[5];
int w[N], tmp[N];

ll d;
void cdq(int l, int r){
	if(l>=r) return;
	int mid=l+r>>1;
	cdq(l, mid), cdq(mid+1, r);
	
	ll sum=0;
	for(int j=l, i=mid+1, k=l;k<=r;k++)
		if(i>r || j<=mid && w[i]>w[j]) sum++, tmp[k]=w[j++]; // 如果说右区间的指针已经走到边界，或者左区间的b值比较小。
		else d+=sum, tmp[k]=w[i++];
	
	for(int k=l;k<=r;k++) w[k]=tmp[k];  // 完成排序，复制
}

int main(){
	int T; cin>>T;
	while(T--){
		set0(cnt);
		string s; cin>>s;
		for(auto i: s)
			if(i=='A') cnt[1]++;
			else if(i=='T') cnt[2]++;
			else if(i=='N') cnt[3]++;
			else if(i=='O') cnt[4]++;
			
		rep(i,1,4) a[i]=i;
		
		int k=0;
		ll dis=1e18;
		int kt;
		do{
			k++;
			string cur;
			
			vector<int> va, vt, vn, vo;
			rep(i,0,s.size()-1)
				if(s[i]=='A') va.pb(i);
				else if(s[i]=='T') vt.pb(i);
				else if(s[i]=='N') vn.pb(i);
				else if(s[i]=='O') vo.pb(i);
		
			d=0;

			int tot=0;
			rep(i,1,4)
				if(a[i]==1){
					for(auto o: va) w[++tot]=o;
				}
				else if(a[i]==2){
					for(auto o: vt) w[++tot]=o;
				}
				else if(a[i]==3){
					for(auto o: vn) w[++tot]=o;
				}
				else if(a[i]==4){
					for(auto o: vo) w[++tot]=o;
				}
			
			cdq(1, s.size());
			if(d<dis) dis=d, kt=k;
		}while(next_permutation(a+1, a+1+4));
		
		// debug(kt);
		rep(i,1,4) a[i]=i;
		
		int kth=0;
		do{
			kth++;
			if(kth==kt){
				// output ans
				string cur;
				rep(i,1,4) rep(j,1,cnt[a[i]]){
					if(a[i]==1) cur+='A';
					else if(a[i]==2) cur+='T';
					else if(a[i]==3) cur+='N';
					else if(a[i]==4) cur+='O';
				}
				cout<<cur<<endl;
				break;
			}
		}while(next_permutation(a+1, a+1+4));
	}
    return 0;
}