A+B for Polynomials

1002. A+B for Polynomials (25)

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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.

Output

For each test case you should output the sum 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 to 1 decimal place.

Sample Input

2 1 2.4 0 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

本题考察的是两个多项式相加的问题。关键方法是自定义两个int类型的iterator，在连个多项式数组中同时遍历，按顺序一次合并各项。

#include <iomanip>控制流头文件包含方法：

setprecision(n)，设置输出变量的精度。

#include <iostream>#include <vector>#include <iomanip>using namespace std;vector<int> exponents1, exponents2, resultExp;vector<double> coeffients1, coeffients2, resultCoe;int terms1, terms2;int main() {int exponent;double coeffient;cin >> terms1;for (int i = 0; i < terms1; i++){cin >> exponent >> coeffient;exponents1.push_back(exponent);coeffients1.push_back(coeffient);}cin >> terms2;for (int i = 0; i < terms2; i++){cin >> exponent >> coeffient;exponents2.push_back(exponent);coeffients2.push_back(coeffient);}int it1=0, it2=0;for (; it1 < terms1 || it2 < terms2;){if (it1<terms1 && (it2>=terms2 ||  exponents1[it1]>exponents2[it2])){resultExp.push_back(exponents1[it1]);resultCoe.push_back(coeffients1[it1]);it1++;}else if (it2<terms2 && (it1 >= terms1 || exponents1[it1]<exponents2[it2])){resultExp.push_back(exponents2[it2]);resultCoe.push_back(coeffients2[it2]);it2++;}else if (it1<terms1 && it2 < terms2 && exponents1[it1] == exponents2[it2]){coeffient = coeffients1[it1] + coeffients2[it2];if (coeffient!=0){resultExp.push_back(exponents1[it1]);resultCoe.push_back(coeffient);}it1++;it2++;}}cout << fixed;cout << resultCoe.size();if (resultCoe.size()==0){return 0;}cout << " ";for (int i = 0; i < resultCoe.size()-1; i++){cout << resultExp[i] << " ";cout << setprecision(1) << resultCoe[i] << " ";}cout << resultExp.back() << " " << setprecision(1) << resultCoe.back();//cout << endl;//system("pause");return 0;}