【题解】PATA-1009 Product of Polynomials

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PATA-1009 Product of Polynomials 类型 水题

Product of Polynomials(PATA-1009)

题面

This time, you are supposed to find A×B where A and B are two polynomials.

输入

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.

输出

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.

样例输入

1
2
2 1 2.4 0 3.2
2 2 1.5 1 0.5

样例输出

1
3 3 3.6 2 6.0 1 1.6

提示

无

思路

类似【题解】PATA-1002 A+B for Polynomials，修改一下。水题。

代码

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#define Sg(u) ((u)>eps?1:((u)<-eps?-1:0))
#define Abs(u) (Sg(u)>=0?(u):-(u))
#define Ze(u) (!Sg(u))
#define Eq(u,v) (Ze((u)-(v)))
const double eps = 1e-6;
double a[2005], b[2005];

int main()
{
    int an; scanf("%d", &an);
    for(int i=0; i<an; i++){
        int n; scanf("%d", &n);
        scanf("%lf", &b[n]);
    }

    int bn; scanf("%d", &bn);
    for(int i=0; i<bn; i++){
        int n; double x;
        scanf("%d %lf", &n, &x);
        for(int j=0; j<=2000; j++)
            if(Sg(b[j]))
                a[n+j] += x * b[j];
    }

    int ans = 0;
    for(int i=2000; i>=0; i--){
        if(Sg(a[i]))
            ans++;
    }
    printf("%d", ans);

    for(int i=2000; i>=0; i--){
        if(Sg(a[i]))
            printf(" %d %.1lf", i, a[i]);
    }
    printf("\n");

    return 0;
}