1002. A+B for Polynomials (25)

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最新推荐文章于 2022-03-23 20:57:34 发布

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本文介绍了一种解决多项式求和问题的算法，通过读取输入的两个多项式，计算它们的和，并以特定格式输出结果。

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1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

题解:　这是一个多项式求解，　数据量不大，　只有1000，　我们可以把两个多项式的和存储到一个当中，　然后统计指数不为0的个数，　最后从大到小依次输出即可，　注意，　指数为0的不输出．　另外如果指数不为0的个数也为0的话，　只打印一个0即可．

#include <cstdio>
#include <cstring>
#include <queue>

const int N = 1010;
const double EPS = 1e-8;

void read(double *ary) {
	int k, n;
	scanf("%d", &k);
	while (k--) {
		scanf("%d", &n);
		scanf("%lf", ary + n);
	}
}

int main() {
	double Na[N] = {0}, Nb[N] = {0};

	read(Na);
	read(Nb);

	int total = 0;
	for (int i = N - 1; i >= 0; --i) {
		if (!(Nb[i] >= -EPS && Nb[i] <= EPS))
			Na[i] += Nb[i];

		if (!(Na[i] >= -EPS && Na[i] <= EPS)) 
			total++;
	}

	printf("%d", total);

	for (int i = N - 1; i >= 0; --i) {
		if (!(Na[i] >= -EPS && Na[i] <= EPS)) 
			printf(" %d %.1lf", i, Na[i]);
	}

	return 0;
}