Codeforces Global Round 5

Posted 2020-12-02 misuchii

Codeforces Global Round 5

A. Balanced Rating Changes

• 思路：(n)为偶数 则这n个数里面奇数的个数一定为偶数 那么就每次令一个答案为(frac{a_i}{2} - 1) 另一个答案为(frac{a_i}{2} + 1)(注意(a_i)的正负性

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

int n, x;
vector<int> a, ans;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 0; i < n; i ++ ){
        cin >> x;
        a.push_back(x);
    }
    int flag = 1;
    for (int i = 0; i < n; i ++ ){
        if (a[i] & 1){
            if (flag == 1){
                if (a[i] >= 0)
                    ans.push_back(a[i] / 2);
                else
                    ans.push_back(a[i] / 2 - 1);
            }
            else{
                if (a[i] >= 0)
                    ans.push_back(a[i] / 2 + 1);
                else
                    ans.push_back(a[i] / 2);
            }
            flag *= -1;
        }
        else
            ans.push_back(a[i] / 2);
    }
    for (int i = 0; i < n; i ++ )
        cout << ans[i] << "
";
    return 0;
}

B. Balanced Tunnel

• 思路：模拟搞搞就行

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e5 + 10;

int n, ans;
int a[N], b[N], res[N];
bool vis[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i ++ )
        cin >> a[i];
    for (int i = 1; i <= n; i ++ )
        cin >> b[i];
    int j = 1;
    for (int i = 1; i <= n; i ++ ){
        vis[a[i]] = true;
        for (; j <= n; j ++ ){
            if (res[a[i]] || a[i] == b[j])
                break;
            if (!vis[b[j]])
                res[b[j]] = 1;
        }
    }
    for (int i = 1; i <= n; i ++ )
        if (res[i])
            ans ++ ;
    cout << ans << "
";
    return 0;
}

C. Balanced Removals

• 思路：一维可以根据x坐标对所有点排序 然后成对移除 二维的时候把每个x相同的点单独处理 然后剩下y不同的进行排序 继续成对移除 分别都剩下一个数 就一起处理 三维同理

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 5e4 + 10;

int n;

struct Point{
    int x, y, z, id;
}p[N], tmp[N];

inline bool cmp(const Point &p1, const Point &p2){
    if (p1.x != p2.x)
        return p1.x < p2.x;
    if (p1.y != p2.y)
        return p1.y < p2.y;
    return p1.z < p2.z;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i ++ ){
        cin >> p[i].x >> p[i].y >> p[i].z;
        p[i].id = i;
    }
    sort(p + 1, p + n + 1, cmp);
    int j = 0;
    for (int i = 1; i <= n; i ++ ){
        if (p[i].x == p[i + 1].x && p[i].y == p[i + 1].y && i < n){
            cout << p[i].id << " " << p[i + 1].id << "
";
            i ++ ;
        }
        else
            tmp[ ++ j ] = p[i];
    }
    int k = 0;
    for (int i = 1; i <= j; i ++ ){
        if (tmp[i].x == tmp[i + 1].x && i < j){
            cout << tmp[i].id << " " << tmp[i + 1].id << "
";
            i ++ ;
        }
        else
            p[ ++ k ] = tmp[i];
    }
    for (int i = 1; i < k; i += 2 )
        cout << p[i].id << " " << p[i + 1].id << "
";
    return 0;
}

D. Balanced Playlist

• 思路：先特判下(mina * 2)是否(geq maxa) 然后把(a)重复三遍 注意跳出条件

• AC代码

---

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 3e5 + 10;
const int INF = 0x3f3f3f3f;

int n;
int max_a = -INF, min_a = INF;
int a[N];
multiset<int> st;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i ++ ){
        cin >> a[i];
        a[i + 2 * n] = a[i + n] = a[i];
        max_a = max(max_a, a[i]), min_a = min(min_a, a[i]);
    }
    if (min_a * 2 >= max_a){
        for (int i = 1; i < n; i ++ )
            cout << "-1 ";
        cout << "-1
";
        return 0;
    }
    int j = 0;
    for (int i = 1; i <= n; i ++ ){
        for (;;){
            if ((!st.empty() && (*( -- st.end())) > 2 * a[j + 1]) || (j == 3 * n))
                break;
            st.insert(a[ ++ j ]);
        }
        cout << j - i + 1 << " ";
        st.erase(st.lower_bound(a[i]));
    }
    return 0;
}