1002. A+B for Polynomials (25)

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

深深体会到，做题最重要的是，审清楚题意之后，要学会用简单聪明的办法来解决问题。不要盲目写代码，这题就是典型。一开始拿到题，我觉得自己并没有很好地想办法，用计算机的思维来思考解决办法。出去的路上，我突然灵光一现，觉得可以用数组的下标来存放指数，对应的值就是系数，十分钟搞定代码。所以思维很重要！

#include<vector>
#include <sstream>
#include<cmath>
#include<iomanip>
#include<iostream>
#include <ctype.h>
#include <stdlib.h>
#include <algorithm>

using namespace std;
int main()
{
//利用下标表示指数，对应的值即是多项式的系数
int n, m;
float polys[1001] = { 0 };//根据题意，指数不会超过1000
cin >> n;

for (int i = 0; i < n; i++)
{
int exp = 0;
cin >> exp;
float coeff = 0;
cin >> coeff;
polys[exp] += coeff;//将系数存入对应指数的下标中，并加上原有的系数值
}

cin >> m;

for (int i = 0; i < m; i++)
{
int exp = 0;
cin >> exp;
float coeff = 0;
cin >> coeff;
polys[exp] += coeff;
}
int count = 0;
for (int i=1000; i >=0; i--)
{
if (polys[i] != 0)
{
count++;
}
}
cout << count;

for (int i = 1000; i >= 0; i--)
{
if (polys[i] != 0)
{
cout << " " << i << " " << setiosflags(ios::fixed) << setprecision(1)<<polys[i];//注意题目要求小数点精度的问题，保留一位小数点
}
}

return 0;
}