PAT (Advanced Level) Practise 1009. Product of Polynomials (25)

1009. Product of Polynomials (25)

This time, you are supposed to find A*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 N1 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=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 3.22 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

题解：

典型的多项式乘法题目，代码如下。

代码：

#include <cstdio>const int maxn=1005;struct Poly{    int exp;    double cof;}poly[maxn];double ans[2*maxn];int main(){    int n,m;    scanf("%d",&n);    for(int i=0;i<n;i++) scanf("%d%lf",&poly[i].exp,&poly[i].cof);    scanf("%d",&m);    for(int i=0;i<m;i++){        int exp;        double cof;        scanf("%d%lf",&exp,&cof);        for(int j=0;j<n;j++) ans[exp+poly[j].exp]+=cof*poly[j].cof;    }    int cnt=0;    for(int i=0;i<=2000;i++){        if(ans[i]) cnt++;    }    printf("%d",cnt);    for(int i=2000;i>=0;i--){        if(ans[i]) printf(" %d %.1f",i,ans[i]);    }    return 0;}