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?N?1???a?N?1?????N?2???a?N?2?????...?N?K???a?N?K????

where?K?is the number of nonzero terms in the polynomial,?N?i???and?a?N?i?????(i=1,2,?,K) are the exponents and coefficients, respectively. It is given that?1≤K≤10,?0≤N?K??<?<N?2??<N?1??≤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.注意指數最大會到2000， C數組應該開到2001

#include <iostream> #include <cmath> using namespace std;int ka, kb; //A和B中的非零項個數 double A[1001] = {0.0}, B[1001] = {0.0}, C[2001] = {0.0}; int expon; double coe; //指數和系數 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.柳神計算時輸入多項式B時直接進行計算！參考：

#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; }

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