Educational Codeforces Round 123 (Rated for Div. 2) A-E题解

Posted 2023-02-26 ZZXzzx0_0

A. Doors and Keys

思路：
$模拟即可$
时间复杂度：$On$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define y second 
#define x first 
using namespace std;
inline void de(auto x) cout << x << "\\n" ;
const int N = 1e6 + 10 , M = 3010 , mod = 998244353  ;

signed main()

    cf
    
        string s ;
        cin >> s ;
        int a = 0 , b = 0 , c = 0 ;
        int f1 = 0 ;
        for(auto i : s)
        
            if(i == 'r') a ++ ;
            else if(i == 'g') b ++ ;
            else if(i == 'b') c ++ ;
            else if(i == 'R')
            
                if(a >= 1) a -- ;
                else f1 = 1 ;
            
            else if(i == 'G')
            
                if(b >= 1) b -- ;
                else f1 = 1 ;
            
            else
            
                if(c >= 1) c -- ;
                else f1 = 1 ;
            
        
        if(f1) puts("NO") ;
        else puts("YES") ;
    
    return 0;

B. Anti-Fibonacci Permutation

思路：
$构造a数组为1,3,2,4,5.....n$
$next-permutation暴力判断输出即可$
时间复杂度：$O{n}^2$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define y second 
#define x first 
using namespace std;
inline void de(auto x) cout << x << "\\n" ;
const int N = 1e6 + 10 , M = 3010 , mod = 998244353  ;

int a[N] ;
 
signed main()

    cf
    
        int n ;
        cin >> n ;
        fer(i,1,n) a[i] = i ;
        swap(a[2],a[3]) ;
        int k = n ;
        while(next_permutation(a + 1 , a + 1 + n))
        
            int f1 = 0 ;
            fer(i,3,n)
            
                if(a[i] == a[i - 1] + a[i - 2])
                
                    f1 = 1 ;
                
            
            if(!f1)
            
                k -- ;
                for(int i = 1 ; i <= n ; i ++) cout << a[i] << " " ;
                puts("") ;
            
            
            if(!k) break ;
        
    
    return 0;

C. Increase Subarray Sums

思路：
$首先f(i)=max(f(i),f(i-1))$

$注意到n最大只有5000$
$长度为1的子数组有n个$
$长度为2的子数组有n-1个$
$.......$
$长度为n的子数组有1个$
$所以一共有n\ast (n+1)/2个子数组$

$对于f(i)来说,选i个数+=x$
$我们可以选择所有长度>=i的子数组的和+=i\ast x$

$设v[i]数组为所有长度>=i的子数组的和的最大值$
$求所有子数组的和可以用前缀和优化$

$那么答案即为f(i)=max(v[i]+i\ast x,f(i-1))$

时间复杂度：$O{n}^2$

#include <bits/stdc++.h>
#define fer(i,a,b) for(re i = a ; i <= b ; ++ i)
#define der(i,a,b) for(re i = a ; i >= b ; -- i)
#define cf int _; cin>> _; while(_--)
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define y second 
#define x first 
using namespace std;
inline void de(auto x) cout << x << "\\n" ;
const int N = 1e6 + 10 , M = 3010 , mod = 998244353  ;

int n ;
int a[N] , s[N] , v[N] , x ;
 
signed main()

    cf
    
        cin >> n >> x ;
        fer(i,1,n) sf(a[i]) , s[i] = s[i - 1] + a[i] , v[i] = -1e9 ;
        v[0] = 0 ;
        fer(len,1,n)
        
            for(int i = 1 ; i + len - 1 <= n ; i ++)
            
                int j = i + len - 1 ;
                v[len] = max(v[len] , s[j] - s[i - 1]) ;
            
        
        der(i,n-1,0) v[i] = max(v[i + 1] , v[i]) ;
        int res = 0 ;
        fer(i,0,n)
        
            res = max(res , v[i] + i * x) ;
            cout << res << ' ' ;
        
        puts("") ;
    
    
    return 0;

D. Cross Coloring

思路：
$首先注意到题目所说$
$新颜色将应用于每个单元格，无论以上是关于Educational\; Codeforces\; Round\; 123\; (Rated\; for\; Div.\; 2)\; A-E题解的主要内容，如果未能解决你的问题，请参考以下文章$

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