Zhejiang school exercise exercises two data structures 7-2 univariate polynomial multiplication and addition operation (20 minutes)

 

Design of the sum of two functions are univariate polynomial and the product.

Input formats:

Input line 2 minutes, respectively, to each row number of the polynomial given non-zero entries, then descending exponential manner enter a nonzero polynomial coefficient and the exponent (integer not exceeding the absolute value of both 1000). Between numbers separated by a space.

Output formats:

Output in 2 rows, respectively descending exponential manner, and outputs the product polynomial and polynomial coefficients and non-zero entries in the index. Between numbers separated by spaces, but the end can not have extra spaces. Zero polynomial should be output 0 0.

Sample input:

4 3 4 -5 2  6 1  -2 0
3 5 20  -7 4  3 1

Sample output:

15 24 -25 22 30 21 -10 20 -21 8 35 6 -33 5 14 4 -15 3 18 2 -6 1
5 20 -4 4 -5 2 9 1 -2 0

#include <iostream>
#define M 10000
using namespace std;
void print(int data[M]){
    bool ini=true;
    for(int i=M-1;i>=0;i--){
        if(data[i]){
            if(!ini) cout<<" ";
            cout<<data[i]<<" "<<i;
            ini=false;
        }
    }
    if(ini) cout<<"0 0";
    cout<<endl;
}
int main(){
    int a[M]={0};
    int b[M]={0};
    int c[M]={0};
    int d[M]={0};
    int T,coef,expo;
    cin>>T;
    for(int i=0;i<T;i++){
        cin>>coef>>expo;
        a[expo]=coef;
    }
    cin>>T;
    for(int i=0;i<T;i++){
        cin>>coef>>expo;
        b[expo]=coef;
    }
    //乘法
    for(int i=M-1;i>=0;i--){
        if(a[i]){
            for(int j=M-1;j>=0;j--){
                if(b[j]){
                    c[i+j]+=(a[i]*b[j]);
                }
            }
        }
    }
    //加法
    for(int i=M-1;i>=0;i--){
        if(a[i]){
            d[i]+=a[i];
        }
    }
    for(int i=M-1;i>=0;i--){
        if(b[i]){
            d[i]+=b[i];
        }
    }
    print(c);
    print(d);
    system("pause");
    return 0;
}
/** In the debugging process, we should bear in mind is the need to distinguish i and J * /