1002_A+B_for_Polynomials

1002. A+B for Polynomials (25)

时间限制
400 ms
内存限制
32000 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

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解：Hash表索引存储的思路，数组表示。
按精度为小数点后一位输出，一般用printf很方便。在C++里有iomanip头文件支持。具体使用参数见此

#include <iostream>
//#include <iomanip>
#include <stdio.h>
using namespace std;
 
int main()
{
    int m,n;
    int a, b;
    int flag = 0;
    double a_coef[1001], b_coef[1001], sum[1001];
    for(int i = 0; i < 1001; i++)
    {
        a_coef[i] = 0.f;
        b_coef[i] = 0.f;
        sum[i] = 0.f;
    }
    cin >> m;
    for(int i = 0; i < m; i++)
    {
        cin >> a;
        cin >> a_coef[a];
    }
    cin >> n;
    for(int j = 0; j < n; j++)
    {
        cin >> b;
        cin >> b_coef[b];
    }
    for(int k = 0; k < 1001; ++k)
    {
        sum[k] = a_coef[k] + b_coef[k];
        if(sum[k] != 0.f)
            flag++;
    }
    cout<<flag;
    for (int k = 1000; k >= 0; k--)
    {
        if (sum[k] != 0.f)
        {
            printf(" %d %.1f", k, sum[k]);
            //cout << ' '<< k << ' ' << setiosflags(ios::fixed) << setprecision(1) << sum[k];
        }
    }
     return 0;
}