Positive Con Sequences

最新推荐文章于 2021-08-18 22:41:56 发布

原创最新推荐文章于 2021-08-18 22:41:56 发布 · 630 阅读

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题意就是：4个等比或等差的数，缺了一个，你要求出那个数。
思路：对每个位置考虑下

#include"iostream"
#include"cstdlib"
#include"cmath"
using namespace std;
int main()
{  
long long  a,b,c,d,t,x;
while(cin>>a>>b>>c>>d,a!=-1||b!=-1||c!=-1||d!=-1)
{ 
    if(a==-1)
    {   
        x=c/b;
        t=b/x;
        if(t*x==b&&t*x*x==c&&t*x*x*x==d&&t>0&&t<=10000)
           cout<<t<<endl;

           else 
           {
            x=c-b;
            t=b-x;
            if(t+x==b&&t+2*x==c&&t+3*x==d&&t>0&&t<=10000)
            cout<<t<<endl;
            else 
              cout<<"-1"<<endl;
           }
    }

    if(b==-1)
    {
         x=d/c;
        t=c/x;
        if(t/x==a&&t*x==c&&t*x*x==d&&t>0&&t<=10000)
           cout<<t<<endl;
           else 
           {
            x=d-c;
            t=c-x;
            if(t-x==a&&t+x==c&&t+2*x==d&&t>0&&t<=10000)
            cout<<t<<endl;
            else 
              cout<<"-1"<<endl;
           }
    }

    if(c==-1)
    {
         x=b/a;
        t=d/x;
        if(t/x/x==a&&t/x==b&&t*x==d&&t>0&&t<=10000)
           cout<<t<<endl;
           else 
           {
            x=b-a;
            t=b+x;
            if(t-2*x==a&&t-x==b&&t+x==d&&t>0&&t<=10000)
            cout<<t<<endl;
            else 
              cout<<"-1"<<endl;
           }
        }
        if(d==-1)
    {
         x=c/b;
        t=c*x;
        if(t/x/x/x==a&&t/x/x==b&&t/x==c&&t>0&&t<=10000)
           cout<<t<<endl;
           else 
           {
            x=c-b;
            t=c+x;
            if(t-3*x==a&&t-2*x==b&&t-x==c&&t>0&&t<=10000)
            cout<<t<<endl;
            else 
              cout<<"-1"<<endl;
           }

    }
}
    return 0;
}