PAT甲级1009. Product of Polynomials (25)

    xiaoxiao2021-03-25  89

    计算多项式的乘积并输出


    看题目

    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


    要求很简单,输入两个多项式,输出乘积,结果的系数保留一位小数。

    //为了输出方便,本次使用c #include <cstdio> #define MAX 2002 //定义会用到的最大的数组元素个数 int main(void) { int N = 0, exp = 0, count = 0;//分别代表输入总数,幂次,最后输出总数目 double cof = 0.0f;//使用double表示系数 //三个数组,初始化为0,虽然浮点数不能精确表示数但是0还是能的 double pA[MAX / 2] = { 0, }; double pB[MAX / 2] = { 0, }; double result[MAX] = { 0, }; //输入总数 scanf("%d", &N); //输入pA的数据 for (int i = 0; i < N; ++i) { scanf("%d %lf", &exp, &cof); pA[exp] = cof; } //输入pB scanf("%d", &N); for (int i = 0; i < N; ++i) { scanf("%d %lf", &exp, &cof); pB[exp] = cof; } //开始计算乘积 for (int i = 0; i < MAX / 2; ++i) { //pA[i] != 0,能节省点时间就行 for (int j = 0; j < MAX / 2 && pA != 0; ++j) { //参考多项式乘法法则 result[i + j] += pA[i] * pB[j]; } } //记录不为0的result个数 for (int i = 0; i < MAX; ++i) { //这个精确度够了, if (result[i] >= 0.1f || result[i] <= -0.1f) { ++count; } } //输出count printf("%d", count); //输出多项式 for (int i = MAX - 1; i >= 0; --i) { if (result[i] != 0.0f) { //精确到小数点后1位 printf(" %d %.1f", i, result[i]); --count; } } //换行,虽然没必要 printf("\n"); //结束 return 0; }

    好好学习,天天向上

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