Codeforces Round #617 (Div. 3) A~D

链接

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e5+10;

int a[N] = {0};

int main(int argc, char const *argv[])
{
    int T;
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--)
    {
        int n,k;
        cin>>n;
        int num=0;
        int flag = 0;
        int temp = 0;
        for (int i = 0; i < n; ++i)
        {
            cin>>a[i];
            if (a[i]&1)
            {
                num++;
            }
            if (i==0)
            {
                temp=a[i];
            }

        }
        if (num<n&&num>=1||num==n&&n&1)
        {
                cout<<"YES"<<endl;
        }
        else
                cout<<"NO"<<endl;
    }
    return 0;
}

链接

每次不花个位数上的钱就可以

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;



int main(int argc, char const *argv[])
{
    int T;
    ios::sync_with_stdio(false);
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        ll ans = 0;
        int temp = n;
        int k = 1;
        int t = 1;
        while(n>=10)
        {
            ans += (n/10)*10; //花更多
            n = n-(n/10)*10+n/10;

        }
        ans += n;
        cout<<ans<<endl;
    }
    return 0;
}

链接

消除一段而不影响终点，相当于删除的这一段走的其实是一个圈
建立map映射点和这段序列的下标即可

#include<bits/stdc++.h>

using namespace std;

const int N = 2e5+10;

char a[N];

map<pair<int,int>,int> mp;
 

int main(int argc, char const *argv[])
{
    int T;
    cin>>T;
    while(T--)
    {
        int n = 0;
        cin>>n;
        cin>>a;
        int ansl=0,ansr=0;
        int t=0,m=1e9+10;
        mp.clear();
        pair<int,int> temp;
        //mp[temp]=1;
        for (int i = 0; i < n; ++i)
        {
            if (a[i]=='L')
            {
                temp.first -= 1;
            }
            if (a[i]=='R')
            {
                temp.first = temp.first+1;
            }
            if (a[i]=='U')
            {
                temp.second = temp.second+1;
            }
            if (a[i]=='D')
            {
                temp.second =temp.second- 1;
            }
            //cout<<temp.first<<" "<<temp.second<<endl;
            if (mp[temp]>0&&i+1-mp[temp]<m||temp==make_pair(0,0)&&i+1-mp[temp]<m)
            {
                ansl = mp[temp]+1;
                ansr = i+1;
                m = i+1-mp[temp];
            }
            mp[temp]=i+1;

        }
        if (m<1e9+10)
        {
            cout<<ansl<<" "<<ansr<<endl;
        }
        else
            cout<<"-1"<<endl;
    }
    return 0;
}

链接

预处理所有妖怪的生命值，算出所需要的跳过的次数，排序后遍历即可

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 2e5+10;

int hp[N] = {0};
vector<int > v;
int main(int argc, char const *argv[])
{
    
    ios::sync_with_stdio(false);
    int n,a,b,k;
    ll ans = 0;
    cin>>n>>a>>b>>k;
    for (int i = 0; i < n; ++i)
    {
        cin>>hp[i];
    }
    for (int i = 0; i < n; ++i)
    {
        if (hp[i]%(a+b)>0&&hp[i]%(a+b)<=a)
        {
            ans++;
        }
        else if(hp[i]%(a+b)==0)
        {
            int gap = b;
            if (gap%a==0)
            {
                v.push_back(gap/a);
            }
            else 
                v.push_back(b/a+1);

        }
        else 
        {
            int temp = (hp[i]%(a+b)-a);
            if (temp%a==0)
            {
                v.push_back(temp/a);
            }
            else
                v.push_back(temp/a+1);
        }
    }
    sort(v.begin(),v.end());
    for (int i = 0; i < v.size(); ++i)
    {
        k-=v[i];
        if (k>=0)
        {
            ans++;
        }
        else
            break;
    }
    cout<<ans<<endl;
    return 0;

}