https://codeforces.com/contest/1974/problem/A

946
A 计算出2*2屏幕所需要的方格 然后计算出剩余出1的格子,相除算出个数不够加一

# include <bits/stdc++.h>
using namespace std;
# define int long long 
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int a,b;
    cin>>a>>b;
    int t=b/2+(b%2==1);
//  cout<<t<<endl;
    a-=t*15-b*4;
    //cout<<a<<endl;
    if(a>=0)
    {
    t+=a/15+(a%15!=0);
    cout<<t;
    }
    else cout<<t;
    cout<<endl;
    }
    return 0;

}

B
本题需要搞懂本体的操作意思，答案操作后变成给出的字符，但是辅助字符串时相同的，对输入用辅助字符串操作后由于字符对称，即可还原成原答案。

# include <bits/stdc++.h>
using namespace std;
# define int long long 
# define x first
# define y second
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int n;
    string s,r="";
    cin>>n>>s;
    map<char,int>ma;
    vector<char>a;
    for(int i=0;i<n;i++)
    {
    ma[s[i]]++;
    }
    for(auto t:ma)
    {
    a.push_back(t.x);
    }
    map<char,char>da;
    int len=a.size();
    for(int i=0;i<len;i++)
    {
    da[a[i]]=a[len-i-1];
    }
    for(int i=0;i<=n-1;i++)
    s[i]=da[s[i]];
    cout<<s<<endl; 
    }
    return 0;
}

C
考察组合数学，分类讨论。对于三元组满足两两相等的条件即对组合答案有贡献，但要注意三个完全相等的情况减去其对于答案的贡献了 用map储存计算答案

# include <bits/stdc++.h>
using namespace std;
# define int long long 
# define x first
# define y second
const int N=2e5+10;
int n;
signed main()
{
int T;
cin>>T;
while(T--)
{
    cin>>n;
    vector<int>a(n+1,0);
    for(int i=1;i<=n;i++)
    cin>>a[i];
    map<pair<int,int>,int>a1;
    map<pair<int,int>,int>a2;
    map<pair<int,int>,int>a3;
    map<array<int,3>,int>b;
    for(int i=2;i<n;i++)
    {
    a1[{a[i-1],a[i]}]++;
    a2[{a[i-1],a[i+1]}]++;
    a3[{a[i],a[i+1]}]++;
    b[{a[i-1],a[i],a[i+1]}]++;
    }
    int  sum=0;
    for(auto t:a1)
    {
    if(t.y>=2)
    sum+=t.y*(t.y-1)/2;
    }
    for(auto t:a2)
    {
    if(t.y>=2)
    sum+=t.y*(t.y-1)/2;
    }
    for(auto t:a3)
    {
    if(t.y>=2)
    sum+=t.y*(t.y-1)/2;
    }
    for(auto &[e,f]:b)
    {
    if(f>=2)
    sum-=3*f*(f-1)/2;
    }
    cout<<sum<<endl;
}
    return 0;

}

960
A
考察博弈内容 如果每个数出现的次数都为偶数那么对于另一个人即可以一直模仿，因此爱丽丝会输

# include <bits/stdc++.h>
using namespace std;
# define int long long
# define x first
# define y second
const int N=52;
int a[N];
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int n;
    cin>>n;
    map<int,int>ma;
    for(int i=1;i<=n;i++)
    cin>>a[i],ma[a[i]]++;
    sort(a+1,a+n+1);
    bool st=0;
    for(auto t:ma)
    if(t.y%2==1)
    st=1;
    if(st)cout<<"YES";
    else cout<<"NO";  
    cout<<endl;
    }

}

B

# include <bits/stdc++.h>
using namespace std;
# define int long long
const int N=1e5+10;
int a[N]={0};
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int n,x,y;
    cin>>n>>x>>y;
    if(x<y)swap(x,y);
    for(int i=y;i<=x;i++)
    a[i]=1;
    a[x+1]=-1;
    a[y-1]=-1;
    //for(int i=1;i<=n;i++)
    //cout<<a[i]<<" ";
    for(int i=y-2;i>0;i--)
    {
    if(a[i+1]==-1)
    a[i]=1;
    else 
    a[i]=-1;
    }
    for(int i=x+2;i<=n;i++)
    {
    if(a[i-1]==1)
    a[i]=-1;
    else 
    a[i]=1;
    }
    for(int i=1;i<=n;i++)
    cout<<a[i]<<" ";
    cout<<endl; 
    }

    return 0;

}