PAT(Advanced)1009 Product of Polynomials (25 分)

1009 Product of Polynomials (25 分)

This time, you are supposed to find $A\times B$ where $A$ and $B$ are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:

$K$ N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ … N​K​​ a​N​K​​​​

where $K$ is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ ($i=1,2,\cdots ,K$) are the exponents and coefficients, respectively. It is given that $1\le K\le 10$, 0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

For each test case you should output the product of $A$ and $B$ in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

题目解读：多项式乘法，输入两个多项式，输出它们乘积的有效项数及结果。

解题思路：仍旧用数组去存储多项式，数组下标表示次数，存一个数组，另一个数组边读边乘，这里要注意把结果的数组开大一点，因为多项式的次数最大为1000，那么如果是两个1000相乘就会是1000+1000=2000，所以结果数组要大一倍。

AC代码

#include <iostream>
using namespace std;
double a[1005],ans[2010];
int main(int argc, const char * argv[]) {
    int tempa,tempb;
    int max = 0;
    int amax = 0;
    int cnt = 0;
    fill(a,a+1005,0);
    fill(ans,ans+1005,0);
    scanf("%d",&tempa);
    for( int i = 0 ; i < tempa ; i ++ ){
        int tempk;
        scanf("%d",&tempk);
        scanf("%lf",&a[tempk]);
        if( tempk > amax )
            amax = tempk;
    }
    scanf("%d",&tempb);
    for( int i = 0 ; i < tempb ; i ++ ){
        int tempk;
        double tempnow;
        scanf("%d %lf",&tempk,&tempnow);
        for( int j = 0 ; j <= amax ; j ++ ){
            if( a[j] == 0 )continue;
            else{
                int ansk = tempk+j;
                double ansnow = tempnow*a[j];
                ans[ansk] += ansnow;
                if( ansk > max )
                    max = ansk;
            }
        }
    }
    for(int i = max ; i >= 0 ; i -- ){
        if( ans[i] != 0 )
            cnt++;
    }
    printf("%d",cnt);
    for( int i = max ; i >= 0 ; i -- ){
        if( ans[i] != 0 ){
            printf(" %d %.1f",i,ans[i]);
            cnt--;
        }
    }
    return 0;
}