A

# include <bits/stdc++.h>
using namespace std;
#define int long long 
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int n;
    cin>>n;
    cout<<n-1<<endl;
    }


}

B

# include <bits/stdc++.h>
using namespace std;
#define int long long 
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    string s;
    cin>>s;
    reverse(s.begin(),s.end());
    for(int i=0;i<s.size();i++)
    {
    if(s[i]=='p')
    s[i]='q';
    else if(s[i]=='q')
    s[i]='p';    
    }
    cout<<s<<endl;
    }
}

C

#include <bits/stdc++.h>
#define int  long long
using namespace std;
const int N=2e5+10;
int m,a,b,c;
void solve()
{
    cin>>m>>a>>b>>c;
    int sum=0,ans=0;
    if(a>=m)
    sum+=m;
    else sum+=a,ans+=m-a;
    if(b>=m)
    sum+=m;
    else sum+=b,ans+=m-b;
    //cout<<sum<<" "<<ans<<endl;
    if(ans)
    {
    sum=sum+min(ans,c);
    }
    cout<<sum<<endl;
}
signed main ()
{
    int T;
    cin >> T;
    while (T --) solve();
    return 0;
}

D
我们可以构造一个排列 使每个数只出现一次也满足要求 先预处理出未出现的数 当重复出现的时候则用未出现的数代替

#include <bits/stdc++.h>
#define int  long long
using namespace std;
const int N=2e5+10;
int n,a[N],b[N];
map<int,int>ma;
set<int>ma1,ma2;
void solve()
{
    ma1.clear(),ma2.clear();
    cin>>n;
    int x=1;
    for(int i=1;i<=n;i++)
    cin>>a[i],ma1.insert(a[i]);
    for(int i=1;i<=n;i++)
    {
    if(ma1.count(i)==0)
    {
    b[x++]=i;
    //cout<<i<<" ";        
    }
    }
    x--;
    int l=1;
    for(int i=1;i<=n;i++)
    {
    if(ma2.count(a[i])==0)
    {
    cout<<a[i]<<" ";
    ma2.insert(a[i]);
    }
    else 
    {
    cout<<b[l++]<<" ";
    }
    }
    cout<<'\n';
}
signed main ()
{
    std::ios::sync_with_stdio(false);
cin.tie(0); 
cout.tie(0);
    int T;
    cin >> T;
    while (T --) solve();
    return 0;
}

E
计算出满足范围的数即可

#include <bits/stdc++.h>
#define int  long long
using namespace std;
const int N=2e5+10;
int k,l1,r1,l2,r2;
void solve()
{
   cin>>k>>l1>>r1>>l2>>r2;
   int ans=0,res=1;
   for(int i=1;i<=32;i++)
   {
   int ll1=(l2+res-1)/res,rr1=r2/res;
   ll1=max(ll1,l1),rr1=min(rr1,r1);
   ans+=max(rr1-ll1+1,(int)0);
   res*=k;
   if(res>1e9)break;
   }
   cout<<ans<<endl;
}
signed main ()
{
    std::ios::sync_with_stdio(false);
cin.tie(0); 
cout.tie(0);
    int T;
    cin >> T;
    while (T --) solve();
    return 0;
}

```