PAT-1002 A+B for Polynomials 解答(C++/Java/python)

1.Description：

This time, you are supposed to find

Notes：

Each case occupies 2 lines. $K$
$1 \leq K \leq 1 0，$

2.Example：

Input:
2 1 2.4 0 3.2
2 2 1.5 1 0.5
output:
3 2 1.5 1 2.9 0 3.2

3.Solutions:

C++ Version:

 1 #include<iostream>
 2 #include<map>
 3 #include<vector>
 4 using namespace std;
 5 
 6 int main() {
 7     map<int, float> polynomial;
 8     for (int i = 0; i < 2; ++i) { //按行存入map里面
 9         int n;
10         scanf("%d", &n); //读取K值
11         for (int i = 0; i < n; ++i) {
12             int exp;
13             float cof;
14             scanf("%d%f", &exp, &cof); //读系数和读值
15             polynomial[exp] += cof; //将对应系数值存入map中
16         }
17     }
18     vector<pair<int, float> > output; //遍历map将值不等0的元组存入数组里
19     for(auto it = polynomial.rbegin(); it != polynomial.rend(); it++) {
20         if(it->second != 0) 
21             output.emplace_back(it->first, it->second);
22     }
23 
24     printf("%d", output.size());
25     for (int i = 0; i < output.size(); ++i) { //格式化输出
26         printf(" %d %.1f", output[i].first, output[i].second);
27     }
28     return 0;
29 }

Java Version:

 1 import java.util.Scanner;
 2 
 3 /**
 4  * Created by SheepCore on 2020-2-26
 5  *
 6  */
 7 public class Main {
 8     public static void main(String[] args) {
 9         final int NUM = 2;
10         final int MAX = 1000;
11         float[] poly = new float[MAX];
12         Scanner scan = new Scanner(System.in);
13         for (int i = 0; i < NUM; i++) { //读取输入数据并按系数存入数组中
14             int n = scan.nextInt();
15             for (int j = 0; j < n; j++) {
16                 int exp = scan.nextInt();
17                 float conf = scan.nextFloat();
18                 poly[exp] += conf;
19             }
20         }
21         int count = 0;
22         int[] out = new int[MAX];
23         for (int i = 0; i < MAX; i++) { //统计个数
24             if (poly[i] != 0.0) {
25                 out[count] = i;
26                 count++;
27             }
28         }
29         System.out.print(count);
30         for(int i = count - 1; i >= 0; i--) { //格式化打印
31             System.out.printf(" %d %.1f", out[i], poly[out[i]]);
32         }
33     }
34 }

Note: 我写的这个Java版本只是部分正确，找了很多遍，没有找出bug，恳请大神指点！

Python Version One:

 1 def solution1():
 2     if __name__ == "__main__":
 3         '''读取两行多项式分割成list'''
 4         a = input().split()
 5         b = input().split()
 6         m = {}
 7 
 8         k1 = eval(a[0])
 9         k2 = eval(b[0])
10 
11         '''将第一行元素按系数存入字典中'''
12         i = 0
13         while i < k1:
14             m[a[2*i+1]] = eval(a[2*i+2])
15             i += 1
16 
17         '''将第二行元素按系数存入字典中'''
18         j = 0
19         while j < k2:
20             if b[2*j+1] in m.keys():
21                 m[b[2*j+1]] += eval(b[2*j+2])
22             else:
23                 m[b[2 * j + 1]] = eval(b[2 * j + 2])
24             j += 1
25         '''统计字典中值为非0的个数'''
26         count = 0
27         for k, v in m.items():
28             if v != 0:
29                 count += 1
30 
31         '''格式化打印'''
32         if count:
33             print(count, end=" ")
34         else:
35             print(count, end="")
36         output = ""
37         if count != 0:
38             for (k, v) in sorted(m.items(), key=lambda x: x[0], reverse=True):
39                 if v != 0:
40                     output += k + " " + str(round(v, 1)) + " "
41             print(output.rstrip(), end='')

Note: 提交的时候不要用函数！

Python Version Two:

 1 def solution2():
 2     output = [0 for i in range(1001)]
 3     a = input().split(" ")
 4     b = input().split(" ")
 5     for i in range(int(a[0])):
 6         output[int(a[i * 2 + 1])] += float(a[i * 2 + 2])
 7     for i in range(int(b[0])):
 8         output[int(b[i * 2 + 1])] += float(b[i * 2 + 2])
 9     count = 0
10     for i in range(1001):
11         if output[i]:
12             count += 1
13     if count:
14         print(count, end=" ")
15     else:
16         print(count)
17     for i in range(1000, -1, -1):
18         if output[i]:
19             count -= 1
20             if count > 0:
21                 print(i, round(output[i], 1), end=" ")
22             else:
23                 print(i, round(output[i], 1))

Note: 提交的时候不要用函数！

4.Summay

如果相加后值为0，不进行输出
size等于0直接输入0（无空格）
输出时只能有一个空格

水滴石穿，笨鸟先飞！：D

PAT-1002 A+B for Polynomials 解答(C++/Java/python)

1.Description：

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

Notes：

2.Example：

3.Solutions:

4.Summay

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