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
#include <iostream> #include <vector> #include <cstdio> using namespace std; int main() { int Na, Nb, tmp1, a_first, b_first; double tmp2; scanf("%d",&Na); vector<double> a(1001); for (int i = 0; i < Na; i++) { scanf("%d %lf", &tmp1, &tmp2); a[tmp1] = tmp2; if (i == 0) a_first = tmp1; } scanf("%d", &Nb); vector<double> b(1001); for (int i = 0; i < Nb; i++) { scanf("%d %lf", &tmp1, &tmp2); b[tmp1] = tmp2; if (i == 0) b_first = tmp1; } int length = a_first + b_first + 1; vector<double> c(length); for (int i = 0; i < length; i++) c[i] = 0; int count = 0; for (int i = 0; i < a_first+1; i++) for (int j = 0; j < b_first+1; j++) { c[i + j] = c[i + j] + a[i] * b[j]; } for (int i = 0; i < length;i++) { if (c[i] != 0) count++; } printf("%d", count); while (length) { if (c[length-1]!=0) printf(" %d %.1f", length-1, c[length-1]); length--; } system("pause"); return 0; }做完才发现多用了一个数组，两个数组就可以了：
for(int i = 0; i < Nb; i++) { scanf("%d %lf", &tmp1, &tmp2); for(int j = 0; j < 1001; j++) b[j + tmp1] += a[j] * tmp2; }