1002. A+B for Polynomials

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 aN1 N2 aN2 ... NK aNK, 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

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

题目内容描述的是两个多项式的加法，原则就是对应指数的系数相加即可，但是从结果看，还需要输出该多项式总共有多少项，其实实质的意思，就是两个数组的加法，所以我们选择两个double类型数组str1和str2，分别保存两组的输入，然后进行对应项相加，也就是str2[i] = str1[i] + str1[i]，结果保存在str2中，其中我们需要判断str2中有多少项非0，统计个数。

#include<stdio.h>
#include<string.h>
using namespace std;

double str1[1002], str2[1002];

int main()
{
freopen("E://input.txt", "r", stdin);
int n, count = 0;
memset(str1, 0, sizeof(str1));
memset(str2, 0, sizeof(str2));

scanf("%d", &n);

for(int i = 0; i < n; i ++)
{
int tmp;
scanf("%d", &tmp);
scanf("%lf", &str1[tmp]);
}

scanf("%d", &n);

for(int i = 0; i < n; i ++)
{
int tmp;
scanf("%d", &tmp);
scanf("%lf", &str2[tmp]);
}

for(int i = 0; i < 1002; i ++)
{
str2[i] += str1[i];
if(str2[i] != 0)
count ++;
}

printf("%d", count);

for(int i = 1001; i >= 0; i --)
{
if(str2[i] != 0)
printf(" %d %.1f", i, str2[i]);
}

printf("\n");

return 0;
}