Educational Codeforces Round 104 Div. 2 A~E

传送门： https://codeforces.com/contest/1487

A-Arena

$输出n-最小数的个数。$

Code

#include "bits/stdc++.h"

using namespace std;

void solve() {
     
    int _; cin >> _;
    while(_--) {
     
        int n; cin >> n;
        map<int, int> mp;
        int mx = 1e9;
        for(int i = 1;i <= n; i++) {
     
            int x; cin >> x;
            mp[x]++;
            mx = min(mx, x);
        }
        cout << n - mp[mx] << endl;
    }
}

signed main() {
     
    solve();
}

B - Cat Cycle

$因为猫B首先呆在1的位置，所以k要减1，最后位置加1即可。$

$当n为奇数时，直接输出(k-1)\%n+1$
$当n为偶数时，因为会有相遇问题。$
$而相遇时间为(n-1)/2，总相遇次数为\frac{(k-1)}{(n-1)/2}。$
$最后位置为(k-1+\frac{(k-1)}{(n-1)/2})\%n+1。$

Code

#include "bits/stdc++.h"

using namespace std;

void solve() {
     
    int _; cin >> _;
    while(_--) {
     
        int n, k; cin >> n >> k;
        k -= 1;
        if(n & 1) {
     
            cout << (k + k / ((n - 1) / 2)) % n + 1 << endl;
        }
        else cout << k % n + 1 << endl;
    }
}

signed main() {
     
    solve();
}

C - Minimum Ties

$当n为奇数时，那么每个队都是\frac{n}{2}的赢和输。所以输出1、-1、1...即可。$

$当n为偶数时，那么每个队都是\frac{n}{2}-1的赢和输还有一场平局。$

$平局场就是1和2，3和4，5和6等等。剩下场都是一半赢一半输。$

Code

#include "bits/stdc++.h"

using namespace std;

void solve() {
     
    int _; cin >> _;
    while(_--) {
     
        int n; cin >> n;
        if(n & 1) {
     
            for(int i = 1;i <= (n - 1) * n / 2; i++) {
     
                if(i & 1) cout << 1 << " ";
                else cout << -1 << " ";
            }
        }
        else {
     
            int t = 0;
            while(n--) {
     
                t++;
                for(int i = 1;i <= n; i++){
     
                    if(i == 1 && (t & 1)) cout << 0 << " ";
                    else {
     
                        if(!(i & 1)) cout << 1 << " ";
                        else cout << -1 << " ";
                    }
                }
            }
        }
        cout << endl;
    }
}

signed main() {
     
    solve();
}

D - Pythagorean Triples

$三个条件：$

• $c=a^2-b$
• $c^2=a^2+b^2$
• $1\le a,b,c\le n$

$问这样(a,b,c)的组数有多少？$

$来化简一下，二式-一式得c^2-c=b^2+b(c+b)(c-b)=b+c$
$继续化简c-b=1b=c-1，带入一式得2c=a^2+1。$

$所以整个条件就变为，a一定要是奇数，并且a\le n，得：$

$$ans=\lceil \frac{\sqrt{2\ast n-1}}{2}\rceil -1$$

$这个减1是因为c=1时，b=0，错误组。$

Code

#include "bits/stdc++.h"

using namespace std;

typedef long long ll;

void solve() {
     
    int _; cin >> _;
    while(_--) {
     
        ll n; cin >> n;
        ll a = sqrt(2 * n - 1);
        cout << a / 2 + (a % 2 == 1) - 1 << endl;
    }
}

signed main() {
     
    solve();
}

E - Cheap Dinner

裸裸地贪心，待补

Code

#include "bits/stdc++.h"

using namespace std;

#define endl "\n"
#define pb push_back

vector<int> a[5], n(5);
vector<pair<int, int>> que, tt;
map<int, set<int>> mp[5];

void solve() {
     
    for (int i = 1; i <= 4; i++) cin >> n[i], a[i].pb(0);
    for (int i = 1; i <= 4; i++) {
     
        for (int j = 1; j <= n[i]; j++) {
     
            int x;
            cin >> x;
            a[i].pb(x);
        }
    }
    for (int i = 1; i <= 3; i++) {
     
        int m;
        cin >> m;
        for (int j = 1; j <= m; j++) {
     
            int x, y;
            cin >> x >> y;
            mp[i][x].insert(y);
        }
    }
    for (int i = 1; i <= n[1]; i++) que.pb({
     a[1][i], i});
    for (int k = 2; k <= 4; k++) {
     
        sort(que.begin(), que.end());
        for (int i = 1; i <= n[k]; i++) {
     
            for (auto j : que) {
     
                int x = j.first, y = j.second;
                if (mp[k - 1][y].find(i) == mp[k - 1][y].end()) {
     
                    tt.pb({
     a[k][i] + x, i});
                    break;
                }
            }
        }
        que = tt;
        tt.clear();
    }
    sort(que.begin(), que.end());
    if (que.empty())
        cout << -1 << endl;
    else
        cout << que[0].first << endl;
}

signed main() {
     
    solve();
}