C. Sakurako’s Field Trip

思路

对于每个位置计算交换以及不交换分别会产生的扰乱度，根据扰乱度的大小决定是否进行交换，最后统计一次答案即可

Solution
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#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
using ll = long long;

void solve() {
    int n;
    cin >> n;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    for (int i = 2; i <= n / 2; i++) {
        int pre = (a[i] == a[i - 1]) + (a[n - i + 1] == a[n - i + 2]);
        int nex = (a[i] == a[n - i + 2]) + (a[n - i + 1] == a[i - 1]);
        if (nex < pre)
            swap(a[i], a[n - i + 1]);
    }
    int ans = 0;
    for (int i = 2; i <= n; i++) {
        if (a[i] == a[i - 1])
            ans++;
    }
    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    // cout << fixed << setprecision(20);
    int tt = 1;
    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}

D. Kousuke’s Assignment

思路

前缀和相同说明有一段为0

Solution
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#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
using ll = long long;

void solve() {
    int n;
    cin >> n;
    vector<ll> v(n);
    vector<ll> pre(n + 1, 0);
    for (int i = 0; i < n; ++i)
        cin >> v[i];
    for (int i = 0; i < n; i++) {
        pre[i + 1] = pre[i] + v[i];
    }
    unordered_map<ll, ll> mp;
    int ans = 0;
    for (int i = 0; i <= n; i++) {
        if (mp[pre[i]]) {
            ans++;
            mp.clear();
            mp[pre[i]] = 1;
        } else {
            mp[pre[i]] = 1;
        }
    }
    cout << ans << '\n';
}

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    // cout << fixed << setprecision(20);
    int tt = 1;
    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}