1009. Product of Polynomials (25)

時(shí)間限制 400 ms 內(nèi)存限制 32000 kB 代碼長(zhǎng)度限制 16000 B 判題程序 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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.

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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 "algorithm" #include <math.h> #include <iomanip> #include"string.h" #include "vector" using namespace std; #define max 3000 struct Poly {int index;//指數(shù)float factor;//系數(shù) };int main() {vector<double> result;int k1,k2;cin >> k1;vector<Poly> p1(k1);for(int i=0;i<k1;i++)cin >> p1[i].index >> p1[i].factor;cin >> k2;vector<Poly> p2(k2);for(int i=0;i<k2;i++)cin >> p2[i].index >> p2[i].factor;result.assign(max+1,0.0);int t=0;for(int i=0;i<k1;i++)for(int j=0;j<k2;j++){Poly p;p.index = p1[i].index+p2[j].index;p.factor = p1[i].factor*p2[j].factor;result[p.index] +=p.factor;}int k=0;for(int i=max;i>=0;i--){if(fabs(result[i])>1e-6){k++;}}cout<<k;for(int i=max;i>=0;i--){if(fabs(result[i])>1e-6){cout<<" "<<i<<" ";cout<<setiosflags(ios::fixed);cout.precision(1);cout<<result[i];}}cout<<endl;return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/luxiao/p/pat1009.html

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