#3238: C++ 一直WA


jeremy83125 (OX)


#include<iostream>
#include<cstdlib>
using namespace std;
int main(int argc,char *argv[])
{
    int a,b,c,d,t,r;
    while(cin>>a,cin>>b,cin>>c,cin>>d)
    {
        if((d-c)==(c-b))
        {
            r=(b-a);
            cout<<a;
            cout<<" "<<b;
            cout<<" "<<c;
            cout<<" "<<d;
            cout<<" "<<(d+r);
        }
        else if((d-c)!=(c-b))
        {
            t=(d/c);
            cout<<a;
            cout<<" "<<b;
            cout<<" "<<c;
            cout<<" "<<d;
            cout<<" "<<(d*t);        
        }       
    }
    system("PAUSE");
    return 0;
}
#3242: Re:C++ 一直WA


leopan0922 (zz)


#include
#include
using namespace std;
int main(int argc,char *argv[])
{
    int a,b,c,d,t,r;
    while(cin>>a,cin>>b,cin>>c,cin>>d)
    {
        if((d-c)==(c-b))
        {
            r=(b-a);
            cout<            cout<<" "<            cout<<" "<            cout<<" "<            cout<<" "<<(d+r);
        }
        else if((d-c)!=(c-b))
        {
            t=(d/c);
            cout<            cout<<" "<            cout<<" "<            cout<<" "<            cout<<" "<<(d*t);        
        }       
    }
    system("PAUSE");
    return 0;
}

第一行是數列的數目t(0 <= t <= 20)。 題目有這行字
#3243: Re:C++ 一直WA


leopan0922 (zz)


#include
#include
using namespace std;
int main(int argc,char *argv[])
{
    int a,b,c,d,t,r;
    while(cin>>a,cin>>b,cin>>c,cin>>d)
    {
        if((d-c)==(c-b))
        {
            r=(b-a);
            cout<            cout<<" "<            cout<<" "<            cout<<" "<            cout<<" "<<(d+r);
        }
        else if((d-c)!=(c-b))
        {
            t=(d/c);
            cout<            cout<<" "<            cout<<" "<            cout<<" "<            cout<<" "<<(d*t);        
        }       
    }
    system("PAUSE");
    return 0;
}

第一行是數列的數目t(0 <= t <= 20)。 題目有這行字


所以改一改變這樣

#include<iostream>
using namespace std;
int main(){
    int n,a,b,c,d;
    while(cin>>n)
    while(n--){
        cin>>a>>b>>c>>d;
        printf("%d %d %d %d %d\n",a,b,c,d,(b-a==c-b)?d+b-a:d*b/a);}
    return 0;}