codeforces ECR169

codeforces ECR 169

A

#include<bits/stdc++.h>
using namespace std;
const int maxn=50;
int a[maxn];
void solve()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    if(n==2)
    {
        if((a[2]-a[1])!=1)
        {
            cout<<"YES\n";
        }
        else
        {
            cout<<"NO\n";
        }
    }   
    else
    {
        cout<<"NO"<<'\n';
    } 
}
int main()
{
    std::ios::sync_with_stdio(0);
    std::cin.tie(0),cout.tie(0);
    int t;
    cin>>t;
    while(t--)solve();
    return 0;
}

B

#include<bits/stdc++.h>
using namespace std;
const int maxn=50;
int a[maxn];
#define   fi     first   
#define   se     second  
void solve()
{
    pair<int ,int>a[2];
    cin>>a[0].fi>>a[0].se;
    cin>>a[1].fi>>a[1].se;
    if(a[0].se<a[1].fi||a[1].se<a[0].fi)
    {
        cout<<1<<'\n';
    }
    else
    {
        int ans=min(a[0].se,a[1].se)-max(a[0].fi,a[1].fi);
        if(a[0].se!=a[1].se)ans++;
        if(a[0].fi!=a[1].fi)ans++;
        cout<<ans<<'\n';
    }
}
int main()
{
    std::ios::sync_with_stdio(0);
    std::cin.tie(0),cout.tie(0);
    int t;
    cin>>t;
    while(t--)solve();
    return 0;
}

C

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxn=2e5+9;
int a[maxn];
void solve()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++)cin>>a[i];
    sort(a+1,a+1+n);
    for(int i=n-1;i>=1;i-=2)
    {
        if(a[i+1]-a[i]<=k)
        {
            k-=a[i+1]-a[i];
            a[i]+=a[i+1]-a[i];
        }
        else
        {
            a[i]+=k;
            k=0;
            break;
        }
    }
    int cnt=1;
    int sum=0;
    for(int i=n;i>=1;i--)
    {
        if(cnt%2==1)
        {
            sum+=a[i];
        }
        else
        {
            sum-=a[i];
        }
        cnt++;
    }
    cout<<sum<<'\n';
}
signed main()
{
    std::ios::sync_with_stdio(0);
    std::cin.tie(0),cout.tie(0);
    int t;
    cin>>t;
    while(t--)solve();
    return 0;
}

D

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 9;
string t[maxn];
string mol[6] = {"BG", "BR", "BY", "GR", "GY", "RY"};
void solve()
{
    int n, q;
    cin >> n >> q;
    map<string, vector<int>> b;
    // BG、BR、BY、GR、GY 或 RY
    b["BG"] = vector<int>(), b["BR"] = vector<int>(), b["BY"] = vector<int>(), b["GR"] = vector<int>(), b["GY"] = vector<int>(), b["RY"] = vector<int>();
    string s;
    for (int i = 1; i <= n; i++)
    {
        cin >> s;
        t[i] = s;
        b[s].push_back(i);
    }
    for (int i = 0; i < 6; i++)
    {
        sort(b[mol[i]].begin(), b[mol[i]].end());
    }
    int x, y;
    for (int i = 1; i <= q; i++)
    {
        cin >> x >> y;
        if (x == y)
        {
            cout << 0 << '\n';
            continue;
        }
        if (t[x][0] == t[y][0] || t[x][0] == t[y][1] || t[x][1] == t[y][0] || t[x][1] == t[y][1])
        {
            int mi = abs(y - x);
            cout << mi << "\n";
        }
        else
        {
            if (x > y)
                swap(x, y);
            int mi = 0x7ffffff;
            for (int _ = 0; _ < 6; _++)
            {
                if ((mol[_].find(t[x][0]) != -1 || mol[_].find(t[x][1]) != -1) && mol[_] != t[x])
                {
                    if (b[mol[_]].size() == 0)
                        continue;
                    if (b[mol[_]][0] > y)
                    {
                        mi = min(mi, abs(2 * b[mol[_]][0] - x - y));
                    }
                    else if (b[mol[_]][b[mol[_]].size() - 1] < x)
                    {
                        mi = min(mi, abs(x + y - 2 * b[mol[_]][b[mol[_]].size() - 1]));
                    }
                    else
                    {
                        mi = min(mi, y - x);
                    }
                }
            }
            cout << ((mi == 0x7ffffff) ? -1 : mi) << '\n';
        }
    }
}
int main()
{
    std::ios::sync_with_stdio(0);
    std::cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

/*
1
1 1
BR 
1 1

*/
posted @ 2024-08-16 12:19  -风间琉璃-  阅读(4)  评论(0编辑  收藏  举报