1009. Product of Polynomials (25)

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

/**    2017.3.6    Donald*///1009. Product of Polynomials (25)/**    Tips：    数组开平方*/#include<cstdio>using namespace std;#define EPS 1e-10#define MAXN 1000001double A[MAXN];double B[MAXN];double Ans[MAXN];void solve(int STARTA, int STARTB){    for(int  i = STARTA; i >= 0; --i){        for(int j = STARTB; j >= 0; --j){            Ans[i + j] += A[i] * B[j];        }    }    int cnt = 0;    for(int i = MAXN - 1; i >= 0; --i){        if(!(Ans[i] >= -EPS && Ans[i] <= EPS))            cnt++;    }    printf("%d", cnt);    for(int i = MAXN - 1; i >= 0; i--)        if(!(Ans[i] >= -EPS && Ans[i] <= EPS))            printf(" %d %.1lf", i, Ans[i]);}int main(void){    int K;    int N;    int startA, startB;    scanf("%d", &K);    scanf("%d", &N);    startA = N;    scanf("%lf", &A[N]);    for(int i = 1; i < K; ++i){        scanf("%d", &N);        scanf("%lf", &A[N]);    }    scanf("%d", &K);    scanf("%d", &N);    startB = N;    scanf("%lf", &B[N]);    for(int i = 1; i < K; ++i){        scanf("%d", &N);        scanf("%lf", &B[N]);    }    solve(startA, startB);    return 0;}