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
多项式乘法
#include<iostream> #include<cstdio> #include<string.h> #include<math.h> #include<algorithm> using namespace std; double s1[1050]; double res[2050]; int main() { int k1,k2; while(~scanf("%d",&k1)) { memset(s1,0,sizeof(s1)); memset(res,0,sizeof(res)); for(int i=1;i<=k1;i++) { int a; double b; scanf("%d%lf",&a,&b); s1[a]=b; } scanf("%d",&k2); for(int i=1;i<=k2;i++) { int a; double b; scanf("%d%lf",&a,&b); for(int j=0;j<1001;j++) { res[j+a]+=s1[j]*b; } } int cnt=0; for(int i=0;i<2002;i++) { if(res[i]!=0) cnt++; } printf("%d",cnt); for(int i=2002;i>=0;i--) { if(res[i]!=0) printf(" %d %.1lf",i,res[i]); } putchar(10); } return 0; }