PAT A1009 Product of Polynomials (25 分)
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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 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???? (,) are the exponents and coefficients, respectively. It is given that 1, 0.
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
#include <stdio.h> #include <stdlib.h> struct poly{ int exp; float a; }; poly p1[11]; double p2[1000010] = { 0 }; const double ep = 1e-4; int main(){ int k1, k2, exp; double a; scanf("%d", &k1); for (int i = 0; i < k1; i++){ scanf("%d %lf", &exp, &a); p1[i].exp = exp; p1[i].a = a; } getchar(); int count = 0; scanf("%d", &k2); for (int i = 0; i < k2; i++){ scanf("%d %lf", &exp, &a); for (int j = 0; j < k1; j++){ double tmp = p1[j].a*a; if (p2[exp + p1[j].exp] == 0 && tmp != 0){ count++; } p2[exp + p1[j].exp] += tmp; if (p2[exp + p1[j].exp] == 0)count--; } } printf("%d", count); for (int i = 1000001; i >= 0; i--){ if (p2[i] != 0.0){ printf(" %d %.1f", i, p2[i]); } } system("pause"); }
注意点:一开始测试点0一直没通过,查了一下是相乘以后再加起来为0,然后以为是精度问题,发现怎么设置ep都不对,后来才知道原来是count计数错了,变为0的count会多加一次。当时在计算时就统计count是怕超时,结果发现多遍历一遍数count不会超时,反而不会错,应该是测试数据没有很大的。
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