Codeforces Round #744 (Div. 3) A-F 题解

Posted 2021-10-18 yueshehanjiang

第一次接近前百纪念

这次的题意就不再赘述了
大概说一下方法

A. Casimir’s String Solitaire

思路：
每次都有B
B的数量等于A和C的和即可
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 all(x) (x).begin(),(x).end()
#define de(x) cout << x << "\\n" 
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 2010 , mod = 1e9 + 7 ;

signed main()
{
    int t ;
    cin >> t ;
    
    while(t--)
    {
        string a ;
        
        int kb = 0 , k = 0 ;
        
        cin >> a ;
        
        for(auto i : a) 
        {
            if(i == 'B') kb ++ ;
            else k ++ ;
        }
        
        if(kb == k) puts("YES") ;
        else puts("NO") ;
    }
    return 0;
}

B. Shifting Sort

思路：
倒着做
每次找到最大的一个数，放到最后即可
注意序列是动态的 不过n很小
直接暴力
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 all(x) (x).begin(),(x).end()
#define de(x) cout << x << "\\n" 
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 60 , M = 2010 , mod = 1e9 + 7 ;
 
int n ;
int a[N] ;
int b[N] ;
int c[N] ;
 
struct ai{
    int l , r , d ;
}q[N] ;
 
signed main()
{
    int t ;
    cin >> t ;
    while(t--)
    {
        cin >> n ;
        
        fer(i,1,n) sf(a[i]) ;
        
        fer(i,1,n) b[i] = a[i] ;
        
        sort(b + 1 , b + 1 + n) ;
        
        int hh = 0 ;
        
        der(i,n,1)
        {
            int k = 0;
            for(int j = i ; j >= 1 ; j --)
            {
                if(a[j] == b[i])
                {
                    k = j ;
                    break ;
                }
            }
            if(k != 0 && k != i)
            {
                q[++ hh] = {k,i,1} ;
                
                fer(i,1,k-1) c[i] = a[i] ;
                fer(i,k,n-1) c[i] = a[i+1] ;
                c[n] = a[k] ;
                memcpy(a,c,sizeof c) ;
            }
        }
        
        cout << hh << '\\n' ;
        fer(i,1,hh) cout << q[i].l << " " << q[i].r << " " << q[i].d << "\\n" ;
        
    }
    return 0;
}

C. Ticks

思路：
模拟一下就行
时间复杂度：$Onmk$

#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 all(x) (x).begin(),(x).end()
#define de(x) cout << x << "\\n" 
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 50 , mod = 1e9 + 7 ;
 
int n, m, k;
char s[M][M];
bool st[M][M];
 
bool check(int x, int y)
{
	if(x < 0 || x >= n || y < 0 || y >= m) return 0 ;
	else return 1 ;
}
 
void get(int x, int y)
{
	int l = 0, r = 0;
 
	while (check(x - l, y - l) && s[x - l][y - l] == '*')
		l ++;
	while (check(x - r, y + r) && s[x - r][y + r] == '*')
		r ++;
 
	l --;
	r --;
	if (min(l, r) >= k) {
		for (int i = 0; i <= min(l, r); i++) {
			st[x - i][y - i] = 1;
			st[x - i][y + i] = 1;
		}
	}
}
 
void solve()
{
    cin >> n >> m >> k;
    
	memset(st, 0, sizeof st);
	
	fer(i,0,n-1)
		scanf("%s", s[i]);
 
	fer(i,0,n-1)
		fer(j,0,m-1)
			if (s[i][j] == '*')
				get(i, j);
 
	bool w = 1;
	fer(i,0,n-1)
		fer(j,0,m-1)
			if (s[i][j] == '*' && st[i][j] == 0)
				w = 0;
 
	if (w)
		puts("YES") ;
	else
		puts("NO") ;
}
signed main()
{
	int t ;
	cin >> t ;
	while (t--) 
	{
		solve() ;
	}
	return 0;
}

D. Productive Meeting

思路：
注意到a[i]的总和是2e5
直接小根堆每次取出最大的2个–即可

其实这题是以前cf的原题
问的是不输出方案的最大值
记录最大值和总和比较一下即可on

其实每次取出最大的2个不一定是对的
（因为给不出证明）
其实我是想用set取出一大一小一定是对的

但是已经写完了优先队列就不想改了
时间复杂度：$Onlogn$

#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 all(x) (x).begin(),(x).end()
#define de(x) cout << x << "\\n" 
#define sf(x) scanf("%lld",&x)
#define pll pair<int,int> 
#define re register int
#define int long long 
#define pb push_back
#define y second 
#define x first 
using namespace std;
const int inf = 0x3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f3f ;
const int N = 1e6 + 10 , M = 2010 , mod = 1e9 + 7 ;

int t ;
int n ;
int a[N] ;
pll ans[N] ;

signed main()
{
    cin >> t ;
    
    while(t--)
    {
        cin >> n ;
    
        int s = 0 ;
        
        priority_queue<pll> q ;
        
        fer(i,1,n)
        {
            sf(a[i]) ;
            if(a[i]) q.push({a[i],i}) ;
        }
        
        int hh = 0 ;
        
        while(q.size() >= 2)
        {
            auto t1 = q.top() ;
            q.pop() ;
            
            auto t2 = q.top() ;
            q.pop() ;
            
            ans[ ++ hh] = {t1.y,t2.y} ;
            
            t1.x -- , t2.x -- ;
            if(t1.x) q.push({t1.x,t1.y}) ;
            if(t2.x) q.push({t2.x,t2.y}) ;
        }
        
        de(hh) ;
        
        fer(i,1,hh)
        {
            cout << ans[i].x << " " << ans[i].y << "\\n" ;
        }
    }
    return 0;
}

E1. Permutation Minimization by Deque

思路：
用duque直接模拟
小的放前面，大的放后面即可
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 all(x) (x).begin(),(x).end()
#define de(x) cout << x << "\\n" 
#define sf(x) scanf("%lld",&x)
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