C. Arithmetic Progression(细节模拟)

https://codeforces.com/problemset/problem/382/C

题意:给定一个序列，问是否可以通过只插入一个数来使得整个序列成为等差数列，求出总共有多少可能的情况，并输出这些数。

思路:

模拟一下发现只有几种情况..然后要考虑各种细节。

其中一个是两个不同的差出现次数相等的时候要用大的差，同时要注意大的差能被2整除且/2后等于小的差

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e5+1000;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar();	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL a[maxn];
map<LL,LL>map1;
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  LL n;cin>>n;
  for(LL i=1;i<=n;i++) cin>>a[i];
  sort(a+1,a+1+n);
  if(n==1){
    cout<<"-1"<<"\n";
  }
  else if(n==2){
     LL cnt=a[2]-a[1];
     if(cnt==0){
        cout<<1<<"\n";
        cout<<a[2]<<"\n";
     }
     else if(cnt%2==0){
        cout<<3<<"\n";
        cout<<a[1]-cnt<<" ";
        cout<<a[1]+cnt/2<<" ";
        cout<<a[2]+cnt<<" ";
     }
     else{
        cout<<2<<"\n";
        cout<<a[1]-cnt<<" ";
        cout<<a[2]+cnt<<" ";
     }
  }
  else{
     for(LL i=1;i<n;i++){
        map1[a[i+1]-a[i]]++;
     }
     if(map1.size()>=3){
        cout<<"0"<<"\n";
     }
     else if(map1.size()==2){
        LL num1=0;LL num2=0;LL cnt1=0;LL cnt2=0;
        for(auto i:map1){
            num1=i.first;cnt1=i.second;
            for(auto j:map1){
                if(i.first==j.first) continue;
                num2=j.first;cnt2=j.second;
            }
            break;
        }
        if(cnt1>=2&&cnt2>=2){cout<<"0"<<"\n";return 0;}
        if(cnt1>=cnt2) swap(num1,num2),swap(cnt1,cnt2);
        if(num1&1||num1/2!=num2){
            cout<<"0"<<"\n";
        }
        else{
            for(LL i=1;i<n;i++){
                if(a[i+1]-a[i]==num1){
                    cout<<"1"<<"\n";
                    cout<<a[i]+num1/2<<"\n";
                    return 0;
                }
            }
        }
     }
     else if(map1.size()==1){
        map<LL,LL>:: iterator it;it=map1.begin();
        LL cnt=(*it).first;
        if(cnt==0){
            cout<<1<<"\n";
            cout<<a[1]<<"\n";
            return 0;
        }
        cout<<2<<"\n";
        cout<<a[1]-cnt<<" "<<a[n]+cnt<<"\n";
     }
  }
return 0;
}

标签：cout,Progression,LL,num2,cnt2,Arithmetic,include,模拟,num1
来源： https://blog.csdn.net/zstuyyyyccccbbbb/article/details/113845616