Educational Codeforces Round 104（A-E题解）

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本文详细解析了EducationalCodeforcesRound104的五道编程竞赛题目，包括思路和解决方案。A题通过查找最小值出现次数求解；B题通过观察规律推导公式得出答案；C题根据不同奇偶性构造比赛结果；D题运用数论方法解决勾股数问题；E题利用优先队列解决分层图的最优化问题。文章展示了如何在有限时间内高效解决各种类型的算法问题。

我昨天打了一下Educational Codeforces Round 104的比赛，场内做出了A,B,C,D,E五题，这场的C,D比以往简单。

A. Arena

思路：直接找最小值出现的次数即可。

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int _;
int n;
int a[110];

void solve(){
    cin >> _;
    while(_--){
        cin >> n;
        int minn = 1e9;
        for(int i = 1; i <= n; i++){
            cin >> a[i];
            minn = min(minn, a[i]);
        }
        int ans = 0;
        for(int i = 1; i <= n; i++){
            if(a[i] == minn){
                ans++;
            }
        }
        cout << n - ans << "\n";
    }
    return ;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    solve();
    return 0;
}

B. Cat Cycle

思路：刚开始被卡了一下，找规律，发现n偶数时候，不会冲突。发现n = 3，会在2，3，4，5，，，，，冲突。n = 5，会在3， 5， 7，，，，冲突。n = 7，会在4，7，10，，冲突，根据规律推公式即可。

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int _;
ll n, k;

void solve(){
    cin >> _;
    while(_--){
        cin >> n >> k;
        if(n % 2 == 0){
            cout << 1 + (k - 1) % n << "\n";
        }
        else{
            cout << 1 + (k - 1 + ((k >= (n/2+1)) ? (1 + (k - (n/2+1))/(n/2)) : (0)))%n << "\n";
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    solve();
    return 0;
}

C. Minimum Ties

思路：构造题。n是奇数，不需要平局，一个人参加n-1场，一半胜利，一半失败。n是偶数，一个人参加n-1场，n-1此时是奇数，所有要有一场平局。

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int _;
int n;

void solve(){
    cin >> _;
    while(_--){
        cin >> n;
        if(n % 2){
            int now = n - 1;
            int tim = now / 2;
            for(int i = 1; i <= n-1; i++){
                for(int j = 1; j <= now; j++){
                    if(j <= tim){
                        cout << "1" << " ";
                    }
                    else{
                        cout << "-1" << " ";
                    }
                }
                now--;
            }
        } 
        else{
            int now = n - 1;
            int tim = now / 2;
            for(int i = 1; i <= n-1; i++){
                for(int j = 1; j <= now; j++){
                    if(j <= tim){
                        cout << "1" << " ";
                    }
                    else if(j == tim + 1){
                        cout << "0" << " ";
                    }
                    else{
                        cout << "-1" << " ";
                    }
                }
                now--;
            }
        }
        cout << "\n";
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    solve();
    return 0;
}

D. Pythagorean Triples

思路：数论题。这个数论比上一场的C简单多了。 $1 < = a, b, c < = n$ 范围内找出 $a^2+b^2=c^2 \&\&a^2-b=c$ 的情况个数。
我们先消除 $a^2$ ,得：
$c^2 - b^2 = c+b$
$c^2-c=b^2+b$
$c * (c - 1) = b (b + 1)$
得出：
$c = b + 1$
可得出：
$a^2=2*b+1$
我们找一下 $a, b, c$ 三个数的范围：
$1 < = a < = n$
$1 < = b < = n$
$1 < = c < = n$
$1 < = b + 1 < = n$
所以：
$1 < = b < = n - 1$
此时：
$3 <= a^2 <= 2*n - 1$
$\sqrt{3} <= a <= \sqrt{2*n-1}$
因为a是整数，我们要找出在 $[3,2∗n−1][\sqrt{3}, \sqrt{2*n-1}]$ 中，找出 $a^2+1)mod2==0$ 的数的个数，就能得到答案。我们可以通过预处理，前缀和的方式，使得每一个询问复杂度达到 $O (1)$ 。

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int _;
const int N = 100010;
ll sum[100010];
ll a[100010];
ll n;

void init(){
    memset(a, 0, sizeof a);
    memset(sum, 0, sizeof sum);
    for(ll i = 2; i <= 100000; i++){
        if(((i*i)-1)%2 == 0){
            a[i] = 1ll;
        }
    }
    for(int i = 1; i <= 100002; i++){
        sum[i] = sum[i-1] + a[i];
    }
}

void solve(){
    cin >> _;
    while(_--){
        cin >> n;
        ll k = sqrt(2ll*n-1ll);
        cout << sum[k] << "\n";
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    init();
    solve();
    return 0;
}

E. Cheap Dinner

思路：这是一个分层图，我们考虑dp[i][j]表示第i层选第j个的最小花费。第i层对j进行规划的时候，要从上一层和j不冲突的dp中选一个最小值，考虑把dp结合优先队列。

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int _;
ll dp[5][200010];
int n1, n2, n3, n4;
ll a[200010];
ll b[200010];
ll c[200010];
ll d[200010];
set<int> vec[5][200010];
struct cmp{
    bool operator () (const pair<ll, int> &a, const pair<ll, int> &b) const{
        return a.first > b.first;
    }
};
priority_queue<pair<ll, int>, vector< pair<ll, int> >, cmp> q;

void solve(){
    cin >> n1 >> n2 >> n3 >> n4;
    for(int i = 1; i <= n1; i++){
        cin >> a[i];
    }
    for(int i = 1; i <= n2; i++){
        cin >> b[i];
    }
    for(int i = 1; i <= n3; i++){
        cin >> c[i];
    }
    for(int i = 1; i <= n4; i++){
        cin >> d[i];
    }
    int m1;
    cin >> m1;
    for(int i = 1; i <= m1; i++){
        int x, y;
        cin >> x >> y;
        vec[2][y].insert(x);
    }
    int m2;
    cin >> m2;
    for(int i = 1; i <= m2; i++){
        int x, y;
        cin >> x >> y;
        vec[3][y].insert(x);
    }
    int m3;
    cin >> m3;
    for(int i = 1; i <= m3; i++){
        int x, y;
        cin >> x >> y;
        vec[4][y].insert(x);
    }
    //cout << m1 << m2 << m3 << "\n";
    memset(dp, 0x3f, sizeof dp);
    for(int i = 1; i <= n1; i++){
        dp[1][i] = a[i];
    }
    for(int tt = 2; tt <= 4; tt++){
        //cout << tt << "\n";
        if(tt == 2){
            while(q.size()){
                q.pop();
            }
            for(int i = 1; i <= n1; i++){
                q.push({dp[1][i], i});
            }
            //cout << q.size() << "\n";
        }   
        else if(tt == 3){
            while(q.size()){
                q.pop();
            }
            for(int i = 1; i <= n2; i++){
                q.push({dp[2][i], i});
            }
        }
        else{
            while(q.size()){
                q.pop();
            }
            for(int i = 1; i <= n3; i++){
                q.push({dp[3][i], i});
            }
        }
        vector< pair<ll, int> > tmp;
        if(tt == 2){
            for(int i = 1; i <= n2; i++){
                tmp.clear();
                while(q.size() != 0 && vec[tt][i].find((q.top().second)) != vec[tt][i].end()){
                    pair<ll, int> now = q.top();
                    q.pop();
                    tmp.push_back(now);
                }
                if(q.size() != 0){
                    pair<ll, int> now = q.top();
                    dp[tt][i] = min(dp[tt][i], now.first + b[i]);
                }
                for(int j = 0; j < tmp.size(); j++){
                    q.push(tmp[j]);
                }
            }
        }
        else if(tt == 3){
            for(int i = 1; i <= n3; i++){
                tmp.clear();
                while(q.size() != 0 &&vec[tt][i].find((q.top().second)) != vec[tt][i].end()){
                    pair<ll, int> now = q.top();
                    q.pop();
                    tmp.push_back(now);
                }
                if(q.size() != 0){
                    pair<ll, int> now = q.top();
                    dp[tt][i] = min(dp[tt][i], now.first + c[i]);
                }
                for(int j = 0; j < tmp.size(); j++){
                    q.push(tmp[j]);
                }
            }
        }
        else{
            for(int i = 1; i <= n4; i++){
                tmp.clear();
                while(q.size() != 0 &&vec[tt][i].find((q.top().second)) != vec[tt][i].end()){
                    pair<ll, int> now = q.top();
                    q.pop();
                    tmp.push_back(now);
                }
                if(q.size() != 0){
                    pair<ll, int> now = q.top();
                    dp[tt][i] = min(dp[tt][i], now.first + d[i]);
                }
                for(int j = 0; j < tmp.size(); j++){
                    q.push(tmp[j]);
                }
            }
        }

    }
    ll ans = 0x3f3f3f3f3f3f3f3f;
    for(int i = 1; i <= n4; i++){
        ans = min(ans, dp[4][i]);
    }
    if(ans >= 0x3f3f3f3f3f3f3f3f){
        cout << "-1" << "\n";
    }
    else{
        cout << ans << "\n";
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    solve();
    return 0;
}