【PAT】PAT A1009 Product of Polynomials【多项式乘法】

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 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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

题目大意

给出两个多项式，求他们的乘积，并输出不为0的项数和各项的幂和底数。

思路

N<=1000，乘积的幂<=2000，因为从0次开始，所以多开一项。因为输入的时候幂按从大到小的顺序输入，所以可以记录最大可能的幂和最小可能的幂，减少遍历。另外在计算的同时计算项数。对于是否为0使用一个极小值来判断，而不是直接判断是否等于0.0，否则会有一组数据无法通过。

代码

#include 
#include 
#include 
using namespace std;
const int maxN = 1001;
// 使用一个极小值来判断项是否为空，否则第一个点过不了
const double err = 1e-7;
double A[maxN];
double R[maxN * 2];
int main() {
    fill(A, A + maxN, 0.0);
    fill(R, R + maxN * 2, 0.0);
    int exp, k1, k2,maxExp = 0, minExp = 0, count = 0;
    double base;
    scanf("%d", &k1);
    for(int i = 0; i < k1; i++){
        scanf("%d %lf", &exp, &base);
        if(i == 0) maxExp = exp;
        else if(i == k1 - 1) minExp = exp;
        A[exp] = base;
    }
    
    scanf("%d", &k2);
    int e;
    for(int i = 0; i < k2; i++){
        scanf("%d %lf", &exp, &base);
        if(i == 0) maxExp += exp;
        else if(i == k2 - 1) minExp += exp;
        for(int i = 0, t = 0;t < k1 && i < maxN; i++){
            if(fabs(A[i]) < err) continue;
            e = exp + i;
            // 原来接近于0，累加后多一项
            if(fabs(R[e]) < err) count++;
            R[e] += A[i] * base;
            // 累加后接近于0，减一项
            if(fabs(R[e]) < err) count--;
            t++;
        }
    }
    
    printf("%d", count);
    for(int i = maxExp; i >= minExp; i--){
        if(fabs(R[i]) < err) continue;
        printf(" %d %.1f", i, R[i]);
    }
    return 0;
}