文章目录
Author: CHEN, Yue
Organization: 浙江大学
Time Limit: 400 ms
Memory Limit: 64 MB
Code Size Limit: 16 KB
A1009 Product of Polynomials (25point(s))
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
Code
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
const int maxn = 1010;
int main(){
double data1[maxn], data2[maxn], data3[2 * maxn]; // 乘积的长度最多翻倍(19、20 408)
int k, exponent, count = 0;
double coefficient;
memset(data1, 0, sizeof(data1)); // data[i]存x^i前的系数
memset(data2, 0, sizeof(data2));
memset(data3, 0, sizeof(data3));
scanf("%d", &k); // 处理第一个多项式的数据
for (int i = 0; i < k; i++){
scanf("%d %lf", &exponent, &coefficient);
data1[exponent] = coefficient;
}
scanf("%d", &k); // 处理第二个多项式的数据
for (int i = 0; i < k; i++){
scanf("%d %lf", &exponent, &coefficient);
data2[exponent] = coefficient;
}
for (int i = 0; i < maxn; i++){ // 对应项相乘,存储结果
for (int j = 0; j < maxn; j++)
data3[i + j] += data1[i] * data2[j]; // 次数相加,系数相乘
}
for (int i = 2 * maxn - 1; i >= 0; i--) { // 计算非零项个数
if (data3[i] != 0.0)
count++;
}
printf("%d", count);
for (int i = 2 * maxn - 1; i >= 0; i--) {
if (data3[i])
printf(" %d %.1f", i, data3[i]);
}
printf("\n");
return 0;
}
Analysis
-已知两个多项式的非零项个数k。依次给出k组数据(形式:次数 该次数前系数)。
-求两个多项式的乘积。输出格式与输入格式相同。
-分别处理第一个多项式和第二个多项式。按照指数相加,系数相乘,将结果存入结果数组。并计算非零项个数。注意系数需要保留一位小数。