Codeforces Round 928 (Div. 4)

A.

// Problem: A. Vlad and the Best of Five
// Contest: Codeforces - Codeforces Round 928 (Div. 4)
// URL: https://codeforces.com/contest/1926/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
int a[N];
void Lan(){
	string s;
	cin>>s;
	int cnta=0;
	for(int i=0;i<s.size();i++){
		cnta+=(s[i]=='A');
	}		
	int cntb=(int)(s.size())-cnta;
	if(cntb>cnta){
		cout<<"B"<<'\n';
	}else{
		cout<<"A"<<'\n';
	}
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int q;
	cin>>q;
	while (q--) {
		Lan();
	}
	return 0;
}

B.

观察图片或者正方形性质

// Problem: B. Vlad and Shapes
// Contest: Codeforces - Codeforces Round 928 (Div. 4)
// URL: https://codeforces.com/contest/1926/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e3 + 9;
int a[N][N];
void Lan(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			char c;
			cin>>c;
			if(c=='1'){
				a[i][j]=1;
			}else{
				a[i][j]=0;
			}
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(a[i][j]==1){
				if(!a[i][j+1] || !a[i+1][j]){
					cout<<"TRIANGLE"<<'\n';
					return;
				}else{
					cout<<"SQUARE"<<'\n';
					return;
				}
			}
		}
	}
	
		
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int q;
	cin>>q;
	while (q--) {
		Lan();
	}
	return 0;
}

C.

前缀和打表，学到了

// Problem: C. Vlad and a Sum of Sum of Digits
// Contest: Codeforces - Codeforces Round 928 (Div. 4)
// URL: https://codeforces.com/contest/1926/problem/C
// Memory Limit: 256 MB
// Time Limit: 500 ms
//
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e5 + 9;
int prefix[N],a[N];
namespace Lan {
    inline string sread() {
        string s=" ";char e=getchar();
        while(e==' '||e=='\n')e=getchar();
        while(e!=' '&&e!='\n')s+=e,e=getchar();
        return s;
    }
    inline void swrite(string s){
        for(char e:s)putchar(e);
        printf("\n");
    }
    inline ll read() {
        ll x=0,y=1;char c=getchar();
        while(!isdigit(c)){if(c=='-')y=-1;c=getchar();}
        while(isdigit(c)){x=(x<<3)+(x<<1)+(c^48);c=getchar();}
        return x*=y;
    }
    inline void write(ll x) {
        if(x<0)x=-x,putchar('-');ll sta[35],top=0;
        do sta[top++]=x%10,x/=10;while(x);
        while(top)putchar(sta[--top]+'0');
    }
}using namespace Lan;
int solve(string x){
    ll res=0;
    for(int i=0;i<x.size();i++){
        res+=(x[i]-'0');
    }
    return res;
}
void lan(){
    int n;
    n=read();
   	write(prefix[n]);
   	putchar('\n');
}
void init(){
    for(int i=1;i<=200000;i++){//打表前缀和处理
        a[i]=solve(to_string(i));
    }
    for(int i=1;i<=200000;i++){
        prefix[i]=prefix[i-1]+a[i];
    }
}
int main() {
    // ios::sync_with_stdio(false);
    // cin.tie(0),cout.tie(0);
    init();
    int q;
    q=read();
    while (q--) {
        lan();
    }
    return 0;
}

D.

题意理解发现是xor操作

// Problem: D. Vlad and Division
// Contest: Codeforces - Codeforces Round 928 (Div. 4)
// URL: https://codeforces.com/contest/1926/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
int a[N];
void Lan(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	map<int,int> mp;
	ll ans=0;
	for(int i=1;i<=n;i++){
		if(mp[a[i]^INT_MAX]){//如果有配对的就可以在一组
			mp[a[i]^INT_MAX]--;
		}else{//没有无法配对
			mp[a[i]]++;
			ans++;
		}
	}
	cout<<ans<<'\n';
	
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int q;
	cin>>q;
	while (q--) {
		Lan();
	}
	return 0;
}

E.

// Problem: E. Vlad and an Odd Ordering
// Contest: Codeforces - Codeforces Round 928 (Div. 4)
// URL: https://codeforces.com/contest/1926/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define eps 1e-5
#define INF 1e9
using namespace std;
typedef long long ll;
const int N = 2e6 + 9;
int a[N];
void Lan(){
	int n,k;
	cin>>n>>k;
	int t=n;
	int cur=1;
	while(t){
		int pos=t-t/2;
		if(k<=pos){//在这一轮
			cout<<cur*(2*k-1)<<'\n';
			return;
		}
		k-=pos;//超过
		cur*=2;//1248
		t/=2;
	}
	// int cnt=0;进行次数
	// while(k>(n+1)/2){
		// int num=(n+1)/2;
		// k-=num;
		// n/=2;
		// cnt++;
	// }
	// cout<<((2*k-1)<<cnt)<<'\n';//an=a1+(n-1)*d;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	int q;
	cin>>q;
	while (q--) {
		Lan();
	}
	return 0;
}