1002 A+B for Polynomials (25 分)
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 aN1N2aN2 … NKaNK
where K is the number of nonzero terms in the polynomial, Ni and aNi(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10,0≤NK<⋯<N2<N1≤1000.
Output Specification:
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
//我刚开始以为指数也可以为小数……编了一大通,提交上去一堆错
//结果是,指数必为整数,系数必为一位小数
#include <cstdio>
#include <cstring>
#include <cmath>
double coe[1001] = { 0 };//多项式存储,下标为指数,存储数据为系数
int main()
{
int n = 0, m = 0, cnt = 0, i = 0;//cnt为多项式个数
int exp = 0;
double t_coe = 0;
scanf("%d", &n);
for (i = 0; i < n; i++)
{
scanf("%d %lf", &exp, &t_coe);
coe[exp] += t_coe;
}
scanf("%d", &m);
for (i = 0; i < m; i++)
{
scanf("%d %lf", &exp, &t_coe);
coe[exp] += t_coe;
}
i = 0;
while (i <= 1000) if (coe[i++]) cnt++;//统计多项式的个数
printf("%d", cnt);
for (int i = 1000; i>=0; i--)
if (coe[i])
printf(" %d %.1lf", i, coe[i]);//逆序打印,格式为保留一位小数
return 0;
}