1002. A+B for Polynomials (25)

1002. A+B for Polynomials (25)

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

分析：题目是两个多项式相加。有坑点在第一次提交的时候没注意，只拿到了17分。由于采用了指数和系数的表示形式，那么当两个多项式相加的时候，就有可能某一项的系数为0，那么这一项就不用输出了。在做题目的时候想得还是不够全面。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
    double ans[1010];
    int flag[1010];
    memset(flag,0,sizeof(flag));
    int cnt=0;
    for(int input=0;input<2;input++)
    {
        int k;
        scanf("%d",&k);
        {
            for(int i=0;i<k;i++)
            {
                int index;
                double num;
                cin>>index>>num;
                if(flag[index]==0)
                {
                    flag[index]=1;
                    cnt++;
                    ans[index]=num;
                }
                else
                {
                    ans[index]+=num;
                    //判断系数是否为0，若为0去掉这一项
                    if(ans[index]<0.0000000001)
                    {
                        flag[index]=0;
                        cnt--;
                    }
                }
            }
        }
    }
    cout<<cnt;
    int j=1010-1;
    for(;j>=0;j--)
    {
        if(flag[j]==1)
        {
            printf(" %d %.1lf",j,ans[j]);
        }
    }
    cout<<endl;
    return 0;
}