PAT甲级 -- 1009 Product of Polynomials (25 分)

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

25分：

注意：

1.注意指数最大会到2000， C数组应该开到2001

#include <iostream>
#include <cmath>
using namespace std;int ka, kb; //A和B中的非零项个数
double A[1001] = {0.0}, B[1001] = {0.0}, C[2001] = {0.0};
int expon;
double coe;  //指数和系数
int mxa = -1, mxb = -1;int main()
{scanf("%d", &ka);for(int i = 0; i < ka; i++){scanf("%d %lf", &expon, &coe);A[expon] += coe;mxa = max(mxa, expon);}scanf("%d", &kb);for(int i = 0; i < kb; i++){scanf("%d %lf", &expon, &coe);B[expon] += coe;mxb = max(mxb, expon);}for(int i = 0; i <= mxa; i++){for(int j = 0; j <= mxb; j++){if(A[i] != 0 && B[j] != 0){C[i+j] += A[i] * B[j];}}}int ct = 0;for(int i = 2000; i >= 0; i--){if(C[i] != 0) ct++;}printf("%d", ct);for(int i = 2000; i >= 0; i--){if(C[i] != 0) printf(" %d %.1f", i, C[i]);}return 0;
}

2.柳神计算时输入多项式B时直接进行计算！参考：

#include <iostream>
using namespace std;
int main() {int n1, n2, a, cnt = 0;scanf("%d", &n1);double b, arr[1001] = {0.0}, ans[2001] = {0.0};for(int i = 0; i < n1; i++) {scanf("%d %lf", &a, &b);arr[a] = b;}scanf("%d", &n2);for(int i = 0; i < n2; i++) {scanf("%d %lf", &a, &b);for(int j = 0; j < 1001; j++)ans[j + a] += arr[j] * b;}for(int i = 2000; i >= 0; i--)if(ans[i] != 0.0) cnt++;printf("%d", cnt);for(int i = 2000; i >= 0; i--)if(ans[i] != 0.0)printf(" %d %.1f", i, ans[i]);return 0;
}