2024-2025 ICPC, NERC, Southern and Volga Russian Regional Contest

自己vp了一下这一场，赛时7题，比较简单，但是有几题也是卡了蛮久。
都是思维题。

C

感觉结论比较显然但是实现上被卡住了。
用map没过，重构的时候把多个数压缩成一个数处理ac了，对拍发现是因为循环逻辑导致错误了。。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define int long long
#define lowbit(x) (x & (-x))
const int N = 2e5 + 10;
int n, ans;
void solve() {
    cin >> n;
    map<int, int>mp;
    for (int i = 1; i <= n; i++) {
        int x; cin >> x;
        mp[x]++;
    }
    int j = 0;
    int b[5] = {0, 0, 0, 0, 0}, k = 0;
    for (auto &i : mp) {
        // cout << i.first << ' ' << i.second << endl;
        if (k < 4){
        while(k + 1 < 5 && i.second >= 2) {
            i.second -= 2;
            b[++k] = i.first;
        }}
        if (k == 4) {
            while(i.second >= 2) {
                i.second -= 2;
                b[3] = i.first;
                if (b[3] >= b[4]) swap(b[3], b[4]);
            }
            continue;
        }
    }
    if (k < 4) cout << "NO" << endl;
    else {
        cout << "YES" << endl;
        cout << b[1] << ' ' << b[2] << ' ' << b[1] << ' ' << b[4] << ' ';
        cout << b[3] << ' ' << b[2] << ' ' << b[3] << ' ' << b[4] << endl;
    }
}
signed main() {
    int T = 1;
    cin >> T;
    while (T--) solve();
}

G

交互题
因为只需要知道一个数，去考虑1后面接的到底是什么数，所以问11，1，10三个，就可以知道1后面到底有没有每个都接着数字

B

之前见过一些类似题所以思路想的比较快
结论1是操作n次可以使全部值-1，也就是说答案具有单调性可以二分
结论2是说暴力check的次数不会超过 log(Amax) 次因为每次可以操作的数都会至少减半
所以可以直接写（在结论2上面卡了很久没搞对）

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define int long long 
#define lowbit(x) (x & (-x))
const int N = 2e5 + 10;
int n, ans, k, a[N], idmx, b[N];
LL sum = 0;
bool jud(int mid) {
    LL op = 1ll * mid * n + k;
    int avg = (sum - op) / n;
    LL cnt = 0;
    for (int i = 1; i <= n; i++) b[i] = a[i];
    for (int i = 1; ; i++) {
        if (op == 0) break;
        int j = ((i - 1) % n) + 1, tmp = (i % n) + 1;
        if (b[j] > avg) {
            b[tmp] = b[tmp] + (b[j] - avg + 1) / 2;
            op -= (b[j] - avg + 1) / 2;
            b[j] -= (b[j] - avg + 1) / 2 * 2;
        }
        if (b[tmp] <= avg && i > n) break;
        if (op < 0) return 0;
    }
    for (int i = 1; i <= n; i++) 
        if (b[i] != avg) return 0;
    return 1;
}
void solve() {
    cin >> n;
    sum = 0;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sum += a[i];
    }
    bool f = 1;
    for (int i = 1; i < n; i++) {
        if (a[i] != a[i + 1]) {
            f = 0; break;
        }
    }
    if (f) {
        cout << 0 << endl;
        return;
    }
    k = sum % n;
    int l = 0, r = sum / n , mid = (l + r) / 2;
    while (l < r) {
        if (jud(mid)) r = mid;
        else l = mid + 1;
        mid = (l + r) / 2;
    }
    // cout << sum << ' ' << n <<  << endl;
    if (jud(mid) == 0) cout << -1 << endl;
    else cout << 1ll * mid * n + k << endl;
}
signed main() {
    int T = 1;
    cin >> T;
    while (T--) {
        solve();
    }
}

补题部分待续