• 题目
• 样例
• 思路
• 完整代码

题目

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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.

样例

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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

思路

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第一次用结构体写的，死活满分不了。
这次AC用的数组。
坑的点有：
1.系数为0的时候，就不输出了。注意空格处理。卡很久。
2.注意输出时候的长度，不是n1+n2-1,而是1000.这点卡了好久。

完整代码

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#include <bits/stdc++.h>


using namespace std;
int main() {
	double s[1001];
	int n1, n2;
	int cnt = 0;
	int index;
	double temp;
	cin >> n1;
	memset(s, 0, sizeof(s));
	for(int i = 0; i < n1; i++){
		cin >> index >> temp;
		s[index] += temp;
	}
	cin >> n2;
	for(int i = 0; i < n2; i++){
		cin >> index >> temp;
		s[index] += temp;
	}
	for(int i = 1000; i >= 0; i--){
		if(s[i] != 0.0){
			cnt++;
		}
	}
	cout << cnt;
	for(int i = 1000; i >= 0; i--){
		if(s[i] != 0.0){
			printf(" %d %.1lf", i, s[i]);
		}
	}
	return 0;
}