PAT 1009 Product of Polynomials
1009. Product of Polynomials (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
#include <iostream> #include <string.h> #include <stdlib.h> #include <algorithm> #include <math.h> #include <stdio.h> #include <string> #include <vector> #include <strstream> #include <map> using namespace std; struct Node { int x; double y; }a[15],b[15]; double tag[2005]; int n1,n2; int main() { scanf("%d",&n1); for(int i=1;i<=n1;i++) scanf("%d%lf",&a[i].x,&a[i].y); scanf("%d",&n2); for(int i=1;i<=n2;i++) scanf("%d%lf",&b[i].x,&b[i].y); memset(tag,0,sizeof(tag)); for(int i=1;i<=n1;i++) { for(int j=1;j<=n2;j++) { tag[a[i].x+b[j].x]+=a[i].y*b[j].y; } } int num=0; for(int i=2000;i>=0;i--) { if(tag[i]!=0) num++; } if(num==0) { printf("%d\n",num); return 0; } else printf("%d ",num); int cnt=0; for(int i=2000;i>=0;i--) { if(tag[i]!=0) { cnt++; if(cnt==num) printf("%d %.1f\n",i,tag[i]); else printf("%d %.1f ",i,tag[i]); } } return 0; }