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 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

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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;}