PAT A1002 A+B for Polynomials

题目描述：

　　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 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.

　　Output Specification:
　　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 3.2
　　2 2 1.5 1 0.5

　　Sample Output:
　　3 2 1.5 1 2.9 0 3.2

参考代码：

 1 /****************************************************
 2 PAT A1002 A+B for Polynomials
 3 ****************************************************/
 4 #include <iostream>
 5 #include <iomanip>
 6 #include <cmath>
 7 #include <vector>
 8
 9 using namespace std;
10
11 struct term {
12     double exponent = 0;
13     double coefficient = 0;
14 };
15
16 int main() {
17     int K1 = 0, K2 = 0;
18     double tempExp = 0, tempCoe = 0;
19
20     cin >> K1;
21     vector<term> func1(K1);
22     for (int i = 0; i < K1; ++i) {
23         cin >> tempExp >> tempCoe;
24         func1[i].exponent = tempExp;
25         func1[i].coefficient = tempCoe;
26     }
27
28     cin >> K2;
29     vector<term> func2(K2);
30     for (int i = 0; i < K2; ++i) {
31         cin >> tempExp >> tempCoe;
32         func2[i].exponent = tempExp;
33         func2[i].coefficient = tempCoe;
34     }
35
36     vector<term> func3(K1 + K2);
37     int i = 0, j = 0, k = 0;
38     for (; i < func1.size() && j < func2.size(); ) {
39         if (func1[i].exponent == func2[j].exponent) {
40             func3[k].exponent = func1[i].exponent;
41             func3[k++].coefficient = func1[i++].coefficient + func2[j++].coefficient;
42         }
43         else if (func1[i].exponent > func2[j].exponent) {
44             func3[k].exponent = func1[i].exponent;
45             func3[k++].coefficient = func1[i++].coefficient;
46         }
47         else {
48             func3[k].exponent = func2[j].exponent;
49             func3[k++].coefficient = func2[j++].coefficient;
50         }
51
52         if (func3[k - 1].coefficient == 0 || fabs(func3[k - 1].coefficient) < 0.1) {
53             func3.erase(func3.begin() + k - 1);
54             --k;
55         }
56     }
57     func3.resize(k);
58     func3.insert(func3.end(), func1.begin() + i, func1.end());
59     func3.insert(func3.end(), func2.begin() + j, func2.end());
60
61     func3.size() == 0 ? cout << ‘0‘ : cout << func3.size() << ‘ ‘;
62     for (int i = 0; i < func3.size(); ++i) {
63         cout << setiosflags(ios::fixed) << setprecision(0) << func3[i].exponent << ‘ ‘;
64         cout << setiosflags(ios::fixed) << setprecision(1) << func3[i].coefficient;
65         if (i != func3.size() - 1) cout << ‘ ‘;
66     }
67
68     return 0;
69 }

注意事项：

　　1：用空间换取时间也是一种减少代码运行时间的方法。

原文地址：https://www.cnblogs.com/mrdragon/p/11396041.html