pat乙级-1010 一元多项式求导 (25)

#include <cmath>
#define N 1001
using namespace std;

int main()
{
    int a[N],b[N],i,k;
    char p;
    for(i=0; p!='n';++i)
    {
        cin>>a[i]>>b[i];
        if(abs(a[i])>1000||b[i]>1000||b[i]<0||a[i]==0&&b[i]!=0)
            return 0;
            if(b[0]==0)
            {
                cout<<"0 0";
                return 0;
            }
        p=cin.get();
    }
    k=i;
    for(i=0;i<=k;++i)
    {

        if(b[i]==0)
            return 0;
        if(i<=k-1&&b[i]<=b[i+1])
            return 0;
    if(a[i]!=0)
        {
            cout<<b[i]*a[i]<<" "<<b[i]-1;
    if(i<k&&b[i+1]!=0)
        cout<<" ";

        }
    }


    return 0;
}