HDU 2020 Multi-University Training Contest 6

Little rabbit's equation
直接枚举即可。

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;
string a,a1,a2,a3;
int u=2,num=1;
char s;
long long int cal(string t,int k)
{
    long long int ans=0;
    for(int i=0;i<t.size();i++)
    {
        if(isdigit(t[i]))
        ans=ans*k+t[i]-'0';
        else
        ans=ans*k+t[i]-'A'+10;
    }
    return ans;
}
int main()
{
    int flag=1;
    while(cin>>a)
    {
        a1.clear();
        a2.clear();
        a3.clear();
        u=2;
        flag=1;
        num=1;
        for(int i=0;i<a.size();i++){
        if(!isdigit(a[i])&&!isalpha(a[i]))
        {
            if(a[i]=='=')
            num++;
            else
            {
                s=a[i];
                num++;
            }
        }
        else if(isdigit(a[i])){
            u=max(u,a[i]-'0'+1);
            if(num==1)
            a1+=a[i];
            else if(num==2)
            a2+=a[i];
            else
            a3+=a[i];
        }
        else
        {
            u=max(u,a[i]-'A'+11);
            if(num==1)
            a1+=a[i];
            else if(num==2)
            a2+=a[i];
            else
            a3+=a[i];
            }    
    }
//    cout<<a1<<endl<<a2<<endl<<a3<<endl;
  for(int i=u;i<=16;i++)
  {
      long long int c1=cal(a1,i),c2=cal(a2,i),c3=cal(a3,i);
      //cout<<c1<<" "<<c2<<" "<<c3<<" "<<endl;
      if(s=='+')
      {
          if(c1+c2==c3)
           {
               cout<<i<<endl;
               flag=0;
               break;
           }
      }
      else if(s=='-')
      {
          if(c1-c2==c3)
          {
              cout<<i<<endl;
              flag=0;
              break;
          }
      }
      else if(s=='*')
     {
         if(c1*c2==c3)
         {
             cout<<i<<endl;
             flag=0;
             break;
         }
     }
     else 
     {
         if(c2==0)
         continue;
         else if(c1%c2!=0)
         continue;
         else if(c1/c2==c3)
         {
             cout<<i<<endl;
             flag=0;
             break;
         }
     }
  }
        if(flag)
        cout<<-1<<endl;
    }
     getchar(); getchar();
     return 0;
}

Divisibility
只要b%x==1即可成立

#include<iostream>
using namespace std;
int main()
{
    int t;
        cin>>t;
        for(int i=0;i<t;i++)
        {
            long long int b,x;
            cin>>b>>x;
            if(b%x==1)
            cout<<'T'<<endl;
            else
            cout<<'F'<<endl;
         } 
} 

Road To The 3rd Building
题目就是求
写出一部分分母相同的式子，可以发现规律：k和n−k+1的式子是相同的，且两边对称，由两边向中间 的系数逐渐增加，当k在一定范围内增加时 的最大系数逐渐增加。

#include <bits/stdc++.h>
#define MAXN 200005
using namespace std;
const long long M=1000000007;
int T,n,a[MAXN];
long long inv[MAXN],sinv[MAXN],ans,C;

inline long long fpow(long long a,long long b)
{
    a%=M;long long r=1;
    for (;b;b>>=1,(a*=a)%=M) if (b&1) (r*=a)%=M;
    return r;
}

void init(int n)
{
    inv[1]=1;
    for (int i=2;i<=n;i++) inv[i]=(M-M/i)*inv[M%i]%M;
    for (int i=1;i<=n;i++) sinv[i]=(sinv[i-1]+inv[i])%M;
    return ;
}

inline void Solve()
{
    scanf("%d",&n);C=ans=0;
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    for (int i=1,p=n>>1;i<=p;i++)
    {
        long long cov=i;
        cov+=i*(sinv[n-i]-sinv[i]+M)%M;cov%=M;
        (C+=i*inv[n-i+1]%M)%=M;(cov+=C)%=M;
        (ans+=a[i]*cov%M)%=M;(ans+=a[n+1-i]*cov%M)%=M;
    }
    if (n&1)
    {
        long long cov=0;
        for (int i=1;i<=n;i++) (cov+=min(i,n+1-i)*inv[i]%M)%=M;
        (ans+=cov*a[(n>>1)+1]%M)%=M;
    }
    printf("%lld\n",ans*fpow(1LL*n*(n+1)/2,M-2)%M);
    return ;
}

int main()
{
    init(200000);
    scanf("%d",&T);
    while (T--) Solve();
    return 0;
}