1009 Product of Polynomials (25)

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本文介绍了一种解决多项式乘法问题的算法实现方法，通过遍历输入的两个多项式，利用双重循环完成指数相加和系数相乘的操作，最终输出结果遵循特定格式。

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 3.2
2 2 1.5 1 0.5
Sample Output

3 3 3.6 2 6.0 1 1.6

解题思路：
1.多项式乘法，和加法类似，最简单明了的方法就是依次读入两个多项式，用两个for循环遍历其中每个式子，指数相加，系数相乘，得到新的多项式。
2.先遍历多项式中的非零项统计个数，再从后向前输出非零项的指数（数组下标），系数（数组的值）。

#include<cstdio>
const int maxn=2001;
double a[maxn]={0.0},b[maxn]={0.0},c[maxn]={0.0};
int main(){
    int k1,k2,exp;
    double cof;
    scanf("%d",&k1);
    for(int i=0;i<k1;i++){
        scanf("%d %lf",&exp,&cof);
        a[exp]=cof;
    }
    scanf("%d",&k2);
    for(int i=0;i<k2;i++){
        scanf("%d %lf",&exp,&cof);
        b[exp]=cof;
    }
    for(int i=0;i<maxn;i++) {
        if(a[i]!=0.0){
            for(int j=0;j<maxn;j++){
                if(b[j]!=0.0){
                    exp=i+j;
                    cof=a[i]*b[j];
                    c[exp]+=cof;
                }
            }
        }
    }
    int cnt=0;
    for(int i=0;i<maxn;i++){
        if(c[i]!=0.0)
            cnt++;
    }
    printf("%d",cnt);
    for(int i=maxn-1;i>=0;i--){
        if(c[i]!=0.0)
            printf(" %d %.1lf",i,c[i]);
    }
    return 0;
}

#include<cstdio>
const int maxn=2001;
double a[maxn];
struct poly{
    int exp;
    double cof;
}poly[1001];
int main(){
    int n1,n2;
    int cut=0;
    //输入第一行 
    scanf("%d",&n1);
    for(int i=0;i<n1;i++){
        scanf("%d %lf",&poly[i].exp ,&poly[i].cof);
    }
    //输入第二行，同时进行A+B操作 
    scanf("%d",&n2);
    int e;
    double k;
    for(int i=0;i<n2;i++){
        scanf("%d %lf",&e,&k);
        for(int j=0;j<n1;j++){
            a[e+poly[j].exp ]+=(k*poly[j].cof );
        }
    }
    for(int i=0;i<2001;i++){
        if(a[i]!=0.0)
            cut++;
    }
    printf("%d",cut);
    for(int i=2000;i>=0;i--){
        if(a[i]!=0.0){
            printf(" %d %.1f",i,a[i]);
        }
    }
    return 0;
}