Educational Codeforces Round 117 (Rated for Div. 2) ABCD

Posted 2022-12-14 陵游gentian

A. Distance
B. Special Permutation
C. Chat Ban
D. X-Magic Pair

A. Distance

思路分析
读懂题意，分情况讨论一下即可

AC代码

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
int T;

int main() 
    cin >> T;
    while (T--) 
        int x, y;
        cin >> x >> y;
        if ((x + y) % 2 != 0) 
            cout << "-1 -1" << endl;
        else 
            if (y > x)
                cout << 0 << " " << (x + y) / 2 << endl;
            else 
                cout << (x + y) / 2 << " " << 0 << endl;
        
    

B. Special Permutation

思路分析
先判断-1的情况，剩下的就比较好处理了

AC代码

#include <bits/stdc++.h>

using namespace std;
typedef pair<int, char> PII;
typedef long long ll;

const int N = 2e5 + 10;

int T;
int n, a, b;
vector<int> l;
vector<int> r;

void solve() 
    l.clear();
    r.clear();
    cin >> n >> a >> b;

    if ((a <= n / 2 && b <= n / 2) || (a > n / 2 && b > n / 2)) 
        cout << -1 << endl;
        return;
    

    if (a > b) 
        if (a - b > 1) 
            cout << -1 << endl;
            return;
         else 
            for (int i = n; i >= 1; i--) 
                cout << i << ' ';
            
            cout << endl;
            return;
        
    

    if (a < b) 
        l.push_back(a);
        r.push_back(b);
        for (int i = n; i > n / 2; i--) 
            if (i != b)
                l.push_back(i);
        

        for (int i = 1; i <= n / 2; i++) 
            if (i != a)
                r.push_back(i);
        

        for (auto item = l.begin(); item != l.end(); item++)
            cout << *item << ' ';
        for (auto item = r.begin(); item != r.end(); item++)
            cout << *item << ' ';
        cout << endl;
    


int main() 
    cin >> T;
    while (T--) 
        solve();
    
    return 0;

C. Chat Ban

思路分析
其实就是三角形的两半，分上下两块三角形进行二分查找即可

AC代码

#include <bits/stdc++.h>
 
using namespace std;
typedef pair<int, char> PII;
typedef long long ll;
 
const int N = 2e5 + 10;
 
int T;
ll x, k;
 
void solve() 
    cin >> k >> x;
 
    if (k * k <= x) 
        cout << 2 * k - 1 << endl;
        return;
    
 
    // 1 到 n 求和
    ll sum = (k * (k + 1)) / 2;
    int ans = 0;
 
    // 三角形上半部分
    if (x <= sum) 
        ll l = 1, r = k;
        ll m;
        while (l <= r) 
            m = l + (r - l) / 2;
            if ((m * (m + 1)) / 2 >= x) 
                ans = m;
                r = m - 1;
             else
                l = m + 1;
        
     else 
        // 三角形下半部分
        x -= sum;
        sum = (k * (k - 1)) / 2;
        ll l = 0, r = k - 1;
        ll m;
        while (l <= r) 
            m = l + (r - l) / 2;
            ll ch = (m * (m + 1)) / 2;
            if (sum - ch < x) 
                ans = m;
                r = m - 1;
             else
                l = m + 1;
        
        ans = 2 * k - ans;
    
    cout << ans << endl;

 
int main() 
    cin >> T;
    while (T--) 
        solve();
    
    return 0;

D. X-Magic Pair

思路分析
这个题是队友给的一个大概思路，当时距离答案已经比较接近了，我回来后改了一下条件就A了，后来想想有点像找规律QAQ

AC代码

#include <bits/stdc++.h>
 
using namespace std;
typedef pair<int, char> PII;
typedef long long ll;
 
const int N = 2e5 + 10;
 
int T;
ll x, a, b;
 
 
int gcd(ll a, ll b, ll x) 
    if (b == 0)
        return 0;
    if (a < x)
        return 0;
    if ((a - x) % b == 0) 
        return 1;
    
    if(gcd(b, a % b, x))
        return 1;
    return 0;

 
int solve() 
    cin >> a >> b >> x;
    if (x > a && x > b) 
        return 0;
    
 
    if (a == b) 
        if (a != x)
            return 0;
        else
            return 1;
     else 
        if(a < b)
            swap(a,b);
        return gcd(a, b, x);
    

 
 
int main() 
    cin >> T;
    while (T--) 
        if (solve())
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    
    return 0;