1002 A+B for Polynomials (25分)（一个字短！）_given n(1≤n≤2×10 5 ) activities {s 1 ,s 2 , ,s n }-优快云博客

本文链接：https://blog.youkuaiyun.com/Ychaoding/article/details/106431973

本文详细介绍了一种解决多项式相加问题的算法实现。通过使用double数组存储多项式的系数，实现了两个多项式的相加运算，并能准确输出结果到小数点后一位。代码示例清晰展示了输入输出规格及关键步骤。

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1002 A+B for Polynomials (25分)

题目

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_N₁… 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<⋯<N₂<N₁≤1000.
Output Specification:
For each test case you should output the sum of AAA and BBB 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

题意

多项式相加

分析

用double数组保存，并相加，最后计数非零项并按照格式输出

代码

#include<iostream>
using namespace std;
double co[1010], d;
int n, a, m = 0, cnt = 0;
int main() {
 	for (int i = 0; i < 2; ++i) {
  		scanf("%d", &n);
  		for (int j = 0; j < n; ++j) {	
			scanf("%d %lf", &a, &d);
   			m = m < a ? a : m;
   			co[a] += d;
  		}
 	}
 	for (int i = m; i >= 0; --i)
  	if (co[i] != 0)++cnt;
 	printf("%d", cnt);
 	for (int i = m; i >= 0; --i)
  		if (co[i] != 0)printf(" %d %.1lf", i, co[i]);
 	return 0;
}