Codeforces Round #679

题目传送门

A. Finding Sasuke

n给定了是偶数，两两配对随便怎么搞都行

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i <= b; i++)
typedef long long ll;
 
ll n;
ll a[110], b[110];
 
void run(int T) {
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    rep(i, 1, n) if(i <= n / 2) cout << a[n - i + 1] << ' ';
    else cout << -a[n - i + 1] <<  " ";
    cout << endl;
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T = 1;
    cin >> T;
    rep(i, 1, T) run(i);
}

B. A New Technique

很显然一个矩阵能确定一个数的其中一维的信息

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i <= b; i++)
typedef long long ll;
 
ll n, m;
pair<ll, ll> ans[250010];
ll mp[510][510];
void run(int T) {
    cin >> n >> m;
    ll f;
    rep(i, 1, n) {
        rep(j, 1, m) {
            cin >> f;
            ans[f].second = j;
        }
    }
    rep(j, 1, m) {
        rep(i, 1, n) {
            cin >> f;
            ans[f].first = i;
        }
    }
    rep(i, 1, n * m) mp[ans[i].first][ans[i].second] = i;
    rep(i, 1, n) {
        rep(j, 1, m) cout << mp[i][j] << ' ';
        cout << '
';
    }
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T = 1;
    cin >> T;
    rep(i, 1, T) run(i);
}

C. Perform Easily

自闭了，对着个假做法敲了好久

实际上把$(b[j] - a[i], j)$提出来排个序，就可以双指针筛了，求个包含$1-j$的最短区间

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i <= b; i++)
typedef long long ll;
 
ll a[7];
ll n, b[100010], ans;
vector<pair<ll, ll>> v;
ll vis[100010], cnt, id;
void run(int T) {
    rep(i, 1, 6) cin >> a[i];
    cin >> n;
    rep(i, 1, n) {
        cin >> b[i];
        rep(j, 1, 6) v.emplace_back(b[i] - a[j], i);
    }
    sort(v.begin(), v.end());
    ans = 1e12;
    for(auto it : v) {
        if(!vis[it.second]) cnt++;
        vis[it.second]++;
        while(cnt == n) {
            ans = min(ans, it.first - v[id].first);
            vis[v[id].second]--;
            if(!vis[v[id++].second]) cnt--;
        }
    }
    cout << ans << endl;
 
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T = 1;
    //cin >> T;
    rep(i, 1, T) run(i);
}

D. Shurikens

栈模拟，再用堆check一下

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i <= b; i++)
typedef long long ll;
 
int n;
stack<int> s;
char f[200010];
int a[200010];
priority_queue<int, vector<int>, greater<int>> q;
void run(int T) {
    cin >> n;
    rep(i, 1, n * 2) {
        cin >> f[i];
        if(f[i] == '-') cin >> a[i]; 
    }
    rep(i, 1, n * 2) {
        if(f[i] == '+') s.push(i);
        else {
            if(s.empty()) {
                cout << "NO
";
                return;
            }
            a[s.top()] = a[i];
            s.pop();
        }
    }
    if(!s.empty()) {
        cout << "NO
";
        return;
    }
    rep(i, 1, n * 2) {
        if(f[i] == '+') q.push(a[i]);
        else {
            if(q.empty() || q.top() != a[i]) {
                cout << "NO
";
                return;
            }
            q.pop();
        }
    }
    cout << "YES
";
    rep(i, 1, n * 2) if(f[i] == '+') cout << a[i] << ' ';
    cout << '
';
 
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T = 1;
    //cin >> T;
    rep(i, 1, T) run(i);
}

E. Solo mid Oracle

判断答案为-1的情况，即$a > b * c$

然后可以推出：消灭最大生命值的时候 一定 在第一次攻击造成的回复之前

然后对于第i次攻击，造成的伤害为$a$，回复的血量为$b*(i - 1)*d$，找到这个$i$即可

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i <= b; i++)
typedef long long ll;


ll a, b, c, d;
void run(int T) {
    cin >> a >> b >> c >> d;
    if(a > c * b) {
        cout << -1 << '
';
        return;
    }

    ll tmp = a / (d * b);
    ll ans = (tmp + 1) * a;
    ans -= tmp * (tmp + 1) / 2 * d * b;
    cout << ans << endl;
}
int main() {
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cout << fixed << setprecision(20);
    int T = 1;
    cin >> T;
    rep(i, 1, T) run(i);
}