1009. Product of Polynomials

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

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
#include<string.h>

int main()
{
    double a[1010] = {}, b[1010] = {}, ans[2010] = {};
    int ka, kb, i, j, x, y, maxa, maxb;
    scanf("%d", &ka);
    scanf("%d", &x);
    scanf("%lf", &a[x]);
    maxa = x;
    for(i = 1; i < ka; i++)
    {
        scanf("%d", &x);
        scanf("%lf", &a[x]);
    }
    scanf("%d", &kb);
    scanf("%d", &y);
    scanf("%lf", &b[y]);
    maxb = y;
    for(i = 1; i < kb; i++)
    {
        scanf("%d", &y);
        scanf("%lf", &b[y]);
    }
    int max = 0;
    if(maxa > maxb)
    {
        for(i = maxa; i >= 0; i--)
        {
            for(j = maxb; j >= 0; j--)
            {
                ans[i + j] += a[i] * b[j];
                if((i + j) > max)
                {
                    max = i + j;
                }
            }
        }
    }
    else
    {
        for(i = maxb; i >= 0; i--)
        {
            for(j = maxa; j >= 0; j--)
            {
                ans[i + j] += b[i] * a[j];
                if((i + j) > max)
                {
                    max = i + j;
                }
            }
        }
    }
    int count = 0;
    for(i = max; i >= 0; i--)
    {
        if(ans[i] != 0)
            count++;
    }
    printf("%d", count);
    for(i = max; i >= 0; i--)
    {
        if(ans[i] != 0)
        {
            printf(" %d %.1f", i, ans[i]);
        }
    }
    printf("
");
    return 0;
}