AtCoder Beginner Contest 346

比赛链接

A - Adjacent Product

思路：

b[i] = a[ i ] * a[ i+1 ]

Code

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

#define int long long
#define all(x) x.begin()+1,x.end()


void solve(){
	int n;
	cin>>n;
	std::vector<int> a(n+1);
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	std::vector<int> b(n);

	for(int i=1;i<n;i++){
		b[i]=a[i]*a[i+1];
	}
	// for(auto x:b+1){
	for(int i=1;i<n;i++){

		cout<<b[i]<<" ";
	}
	cout<<endl;
	return ;
	
}

signed main(){
	int t=1;
	while(t--) solve();
	
	return 0;
}

B - Piano

思路

模拟一下，看数据量构造一个长度位2000的字符串在遍历一下就可以了

Code

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

#define int long long
#define all(x) x.begin()+1,x.end()


void solve(){
	int w,b;
    cin>>w>>b;
    string a="wbwbwwbwbwbw";
    string s=" ";
    while(s.size()<=2000){
        s+=a;

    }
    std::vector<int> n1(2010),n2(2010);
    for(int i=1;i<s.size();i++){
        n1[i]=n1[i-1]+(s[i]=='w');
        n2[i]=n2[i-1]+(s[i]=='b');

    }
    for(int i=1;i<=s.size();i++){
        for(int j=i+1;j<=s.size();j++){
            if((n1[j]-n1[i]==w)&&(n2[j]-n2[i]==b)){
                cout<<"Yes"<<endl;
                return ;
            }
        }
    }
    cout<<"No"<<endl;
    return ;
    

}


signed main(){
	int t=1;


	while(t--){
		solve();
	}
	return 0;
}

C - Σ

思路：

map和数组简单的应用一下就可以了，感觉很好想到

Code

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

#define int long long
#define all(x) x.begin()+1,x.end()


void solve(){
	int n,k;
	cin>>n>>k;
	std::map<int, int> mp;
	std::vector<bool> st(n+1);
	int ans=(k+1)*k/2;
	std::vector<int> a(n+1);
	for(int i=1;i<=n;i++){
		cin>>a[i];
		if(mp[a[i]]==0){
			mp[a[i]]=i;
		}
		if(a[i]<=k&&a[i]>=1){
			if(st[mp[a[i]]]){
				continue;
			}
			else{
				st[mp[a[i]]]=true;
				ans-=a[i];

			}
		}
	}
	cout<<ans<<endl;
	return ;
}

signed main(){
	int t=1;
	while(t--) solve();
	
	return 0;
}

D - Gomamayo Sequence

思路

处理+贪心

Code

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

#define int long long
#define all(x) x.begin()+1,x.end()
const int INF = 1e18;
const int N = 2e5+10;
int fst[N][2];
int lst[N][2];


void solve(){
	int n;
	cin>>n;

	string s;
	cin>>s;
	s=" "+s;//处理一下字符串

	std::vector<int> a(n+1);
	for(int i=1;i<=n;i++){
		cin>>a[i];
		
	}	

	
    for (int i = 1; i <= n; i++) {
        fst[i][0] = fst[i - 1][1] + a[i] * (s[i] != '0');
        fst[i][1] = fst[i - 1][0] + a[i] * (s[i] != '1');
    }
    for (int i = n; i >= 1; i--) {
        lst[i][0] = lst[i + 1][1] + a[i] * (s[i] != '0');
        lst[i][1] = lst[i + 1][0] + a[i] * (s[i] != '1');
    }

    int ans = INF;//记录最小次数

    for (int i = 1; i < n; i++) {
        int now = a[i] * (s[i] != '0') + a[i + 1] * (s[i + 1] != '0');
        now += fst[i - 1][1] + lst[i + 2][1];
        ans = min(ans, now);
    }

    for (int i = 1; i < n; i++) {
        int now = a[i] * (s[i] != '1') + a[i + 1] * (s[i + 1] != '1');
        now += fst[i - 1][0] + lst[i + 2][0];
        ans = min(ans, now);
    }

   	cout << ans << endl;
   	return ;
}


signed main(){
	int t=1;
	while(t--) solve();
	
	return 0;
}