1009 Product of Polynomials (PAT甲级)

这道题我的解法有点复杂化了,其实很直观解就可以。

柳婼解法如下:1009. Product of Polynomials (25)-PAT甲级真题_柳婼的博客-CSDN博客

我的代码如下:

#include 
#include 
const int maxCoef = 2001;

int K, n, tmp, maxx, cnt;
double an;
double coef[maxCoef];
std::map mp;

int main(){
    scanf("%d", &K);
    for(int i = 0; i < K; ++i){
        scanf("%d %lf", &n, &an);
        if(an != 0){
            mp[n] = an;
        }
        if(i == 0){
            maxx = n;
        }
    }
    cnt = 0;
    scanf("%d", &K);
    for(int i = 0; i < K; ++i){
        scanf("%d %lf", &n, &an);
        if(i == 0){
            maxx += n;
        }
        if(an == 0){
            continue;
        }
        for(auto it = mp.begin(); it != mp.end(); ++it){
            tmp = it->first + n;
            if(coef[tmp] == 0){
                cnt++;
            } else if(coef[tmp] + an * it->second == 0){
                cnt--;
            }
            coef[tmp] += an * it->second;
        }
    }
    printf("%d", cnt);
    for(int i = maxx; i >= 0; --i){
        if(coef[i] != 0){
            printf(" %d %.1f", i, coef[i]);
        }
    }
    return 0;
}

题目如下:

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​<⋯

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

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