1009. Product of Polynomials (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue

1、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

2、解析

1）使用数组下标代表指数，数组数值代表系数。

3、代码

#include<iostream>
#define N 2002
using namespace std;
int main() {
    int i, j,l;
    float k;
    float f1[N] = { 0 }, f2[N] = { 0 }, f3[N] = {0};
    int k1, k2;
    float k3;
    int num = 0;
    //读入数据
    cin >> k1;
    for (i = 0;i < k1;++i) {
        scanf("%d %f", &j, &k);
        f1[j] = k;
    }
    cin >> k2;
    for (i = 0;i < k2;++i) {
        scanf("%d %f", &j, &k);
        f2[j] = k;
    }
    //求乘积
    for (i = 0;i < N;++i) {
        //剪枝
        if (f1[i] != 0) {
            for (j = 0;j < N;++j) {
                //剪枝
                if (f2[j] != 0) {
                    l = i + j;
                    k3 = f1[i] * f2[j];
                    f3[l] += k3;
                }
            }
        }
        
    }
    //计算非零个数
    for (i = 0;i < N;++i) {
        if (f3[i] != 0)
            ++num;
    }
    //输出结果
    cout << num;
    for (i = N - 1;i >= 0;--i) {
        if (f3[i] != 0) {
            printf(" %d %.1f", i, f3[i]);
        }
    }
    system("pause");
    return 0;
}