Educational Codeforces Round 158 (Rated for Div. 2)

Educational Codeforces Round 158 (Rated for Div. 2)

A

枚举模拟

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;
int a[N];

void solve()
{
    int n , x;
    cin >> n >> x;
    memset(a , 0 , sizeof a);
    for(int i = 1 ; i <= n ; i ++){
        int t;
        cin >> t;
        a[t] = 1;
    }
    //枚举初始
    for(int i = 0 ; i < 300 ; i ++){
        bool f = true;
        int c = i;
        for(int j = 1 ; j < x ; j ++){
            if(a[j])c = i;
            c --;
            if(c < 0)f=false;
        }
        for(int j = x ; j > 0 ; j --){
            if(a[j])c = i;
            c --;
            if(c < 0)f=false;
        }
        if(f){
            cout << i << endl;
            return ;
        }
    }
}

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

B

当提升时至少要走提升的次数

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;
int a[N];

void solve()
{
    int n;
    cin >> n;
    for(int i = 1 ; i <= n ; i ++){
        cin >> a[i];
    }
    long long ans = 0;
    for(int i = 1 ; i <= n ; i ++){
        if(a[i] > a[i-1]){
            ans += a[i] - a[i-1];
        }
    }
    cout << ans - 1 << endl;
}

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

C

操作不会改变数据的相对性，找出最大和最小让他们接近

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;
int a[N];

void solve()
{
    int n , mx = -2e9 , mi = 2e9;
    cin >> n;
    for(int i = 1 ; i <= n ; i ++){
        cin >> a[i];
        mx = max(mx , a[i]);
        mi = min(mi , a[i]);
    }
    vector<int>ans;
    while(mx != mi){
        int x;
        if (mx % 2 == 0 && mi % 2 == 0) x = 1;
        else if (mx % 2 == 1 && mi % 2 == 1) x = 1;
        else if (mx % 2 == 0 && mi % 2 == 1) x = 1;
        else x = 2;
        mx = (mx + x) / 2;
        mi = (mi + x) / 2;
        ans.push_back(x);
    }
    int len = ans.size();
    cout << len << endl;
    if(len <= n){
        for(auto o : ans)cout << o << " ";
        cout << endl;
    }
}

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

D

前后缀分解左边右边全部击败需要的值
再枚举一开始打哪个

#include <bits/stdc++.h>

using namespace std;

const int N = 3e5 + 10;
int a[N] , l[N] , r[N];

void solve()
{
    int n ;
    cin >> n;
    for(int i = 1 ; i <= n ; i ++){
        cin >> a[i];
    }
    //到i和左边全部击败需要的
    for(int i = 1 ; i <= n ; i ++){
        l[i] = max(l[i-1] , a[i] + n - i);
    }
    for(int i = n ; i >= 1 ; i --){
        r[i] = max(r[i+1] , a[i] + i - 1);
    }
    int ans = 2e9;
    for(int i = 1 ; i <= n ; i ++){
        int cur = max(a[i] , max(l[i - 1] , r[i + 1]));
        ans = min(ans , cur);
    }
    cout << ans << endl;
}

int main()
{
    int T = 1;
    // cin >> T;
    while(T--){
        solve();
    }
    return 0;
}