标签：inpu exponents case algorithm weight mes notice class tac
This time, you are supposed to find A×B where A and B are two polynomials.
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.
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.
2 1 2.4 0 3.2
2 2 1.5 1 0.5
3 3 3.6 2 6.0 1 1.6
注意系数为0的情况
#include<iostream> #include<cstdio> #include<vector> #include<cstring> #include<stack> #include<algorithm> #include<map> #include<string.h> #include<string> #define MAX 1000000 #define ll long long using namespace std; double a[10005],b[10005],c[10005]; double y; int x; int main() { int k; cin>>k; for(int i=0;i<2006;i++) a[i]=b[i]=c[i]=0; for(int i=0;i<k;i++) { cin>>x>>y; a[x]=y; } cin>>k; for(int i=0;i<k;i++) { cin>>x>>y; b[x]=y; } for(int i=0;i<1005;i++) { for(int j=0;j<1005;j++) c[i+j]=c[i+j]+a[i]*b[j];//合并同类项 } int cnt=0; for(int i=0;i<2005;i++) { if(c[i]!=0)//去除多项式系数为0的项 cnt++; } cout<<cnt; for(int i=2005;i>=0;i--) { if(c[i]!=0) printf(" %d %.1lf",i,c[i] ); } cout<<endl; return 0; }
1009 Product of Polynomials (25分) 多项式乘法
标签：inpu exponents case algorithm weight mes notice class tac
原文地址：https://www.cnblogs.com/-citywall123/p/12123007.html
