Educational Codeforces Round 123 VP后反思

Educational Codeforces Round 123

比赛地址

A. Doors and Keys

简单的顺序遍历

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define repd(i, l, r) for (int i = l; i <= r; i++)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define x first
#define y second
#define endl '\n'
using namespace std;
typedef long long LL;
const int N = 1e6 + 10, M = 1010;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

string solve(){
    string s;
    cin >> s;
    bool r_ = 0 , g_ = 0 , b_ = 0;
    rep( i , 0 , s.size() ){
        if( s[i] <= 'z' && s[i] >= 'a' ) {
            if( s[i] == 'r' ) r_ = 1;
            else if( s[i] == 'g' ) g_ = 1;
            else b_ = 1;
        }
        else {
            if( s[i] == 'G' && !g_ ) return "NO";
            if( s[i] == 'B' && !b_ ) return "NO";
            if( s[i] == 'R' && !r_ ) return "NO";
        }
    }
    return "YES";
}

int main()
{
    IOS;
    int t;
    cin >> t;
    while (t--)
    {
        cout << solve() << endl;
    }
    return 0;
}

B. Anti-Fibonacci Permutation

可以根据限制条件写暴搜，也可以根据规律来写。

暴搜

bool st[N];
int c[N];
int t;
int n;

bool dfs(int f)
{
    if(f == n)
    {
        for(int i = 0; i < n; i ++)
            cout << c[i] << " ";
        cout << endl;
        return 1;
    }

    for(int i = 1; i <= n; i ++)
    {
        if(f >= 2)
        {
            if(!st[i] && st[c[f - 1]] && st[c[f - 2]])
            {
                c[f] = i;
                if(c[f - 2] + c[f - 1] != c[f])
                {
                    st[i] = 1;                  
                    if(dfs(f + 1)) return 1;
                    st[i] = 0;
                }
            }
        }
        else
        {           
            if(!st[i])
            {
                c[f] = i;
                st[i] = 1;
                if(dfs(f + 1)) return 1;
                st[i] = 0;
            }
        }
    }

    return 0; // 不要忘了
}

int main(void)
{
    cin >> t;
    while(t --)
    {
        cin >> n;

        for(int i = 1; i <= n; i ++)
        {
            memset(st, 0, sizeof st);
            memset(c, 0, sizeof c);
            c[0] = i;
            st[i] = 1;
            dfs(1);
        }
    }
    return 0;
}

规律

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define repd(i, l, r) for (int i = l; i <= r; i++)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define x first
#define y second
#define endl '\n'
using namespace std;
typedef long long LL;
const int N = 1e6 + 10, M = 1010;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

int a[N];

void solve(){
    
    int n,k=0;
    cin >> n;
    rep( i , 1 , n+1 )
        a[i] = n-k++;
    rep( i , 1 , n+1 ){
        k = 0;
        rep( j , 1 , n+1 ){
            if( i == j ) cout << 1 << ' ';
            else cout << a[++k] << ' ';
        }
        cout << endl;
    }
        
}

int main()
{
    IOS;
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}

C. Increase Subarray Sums

求最大子段和最大。

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define PII pair<int, int>
#define rep(i, l, r) for (int i = l; i < r; i++)
#define repd(i, l, r) for (int i = l; i <= r; i++)
#define rep2(i, l, r) for (int i = l; i * i <= r; i++)
#define rep3(i, l, r) for (LL i = l; i * i * i <= r; i++)
#define Min(a, b) a > b ? b : a
#define Max(a, b) a > b ? a : b
#define endl '\n'
using namespace std;
typedef long long LL;
const int N = 1e6 + 10, M = 1010;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

int a[N],s[N];
int ans[N];

void solve(){
    
    int n,k=0,x;
    cin >> n >> x;
    memset( ans , -0x3f , sizeof ans );
    rep( i , 1 , n+1 ){
        cin >> a[i];
        s[i] = s[i-1]+a[i];
    }
    rep( i , 1 , n+1 ){
        for( int j = 1 ; j+i-1 <= n ; j++ )
            ans[i] = max( ans[i] , s[j+i-1]-s[j-1] ); 
        // 长度为i的区间和的最大值
    }
    
    int Ans = 0;
    rep( k , 0 , n+1 ){
        rep( i , k , n+1 )
            Ans = max( Ans , ans[i]+k*x );
        cout << Ans << ' ';
    }
    cout << endl;
}

int main()
{
    IOS;
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
    return 0;
}