pat 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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

解：有点类似多项式相加，略简单，我第一反应是应该用map，hash应该也行，我用的map，两次输入，根据id值进行判断，相同累加，不同新map[key]=value，要注意的就是当相加之后出现0的情况，对此我们需要额外判断，比如说根据题目要求，如下输入输出应该通过：

Sample Input

2 1 2.4 0 3.2
2 1 -2.4 0 -3.2

Sample Output

0

对于这种情况要特殊考虑，变换一下控制条件。

代码如下：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<map>
using namespace std;
int main()
{
    int cnt1,cnt2=0;
    while(scanf("%d",&cnt1)==1)
    {
        map<int,double> ans;
        int no;
        double value;
        for(int i=0;i<cnt1;i++)
        {
            scanf("%d%lf",&no,&value);
            ans[no]=value;
        }
        scanf("%d",&cnt2);
        for(int i=0;i<cnt2;i++)
        {
            scanf("%d%lf",&no,&value);
            if(ans.find(no)!=ans.end())
            {
                ans[no]+=value;
            }
            else
                ans[no]=value;
        }
        int n=0;
        map<int,double>::iterator q;
        for(q=ans.begin();q!=ans.end();q++)
        {
            if(q->second!=0)
                n++;
        }
        printf("%d",n);
        if(n!=0)
        {
            map<int,double>::reverse_iterator p;
            for(p=ans.rbegin();p!=ans.rend();p++)
            {
                if(p->second!=0)
                {
                    printf(" %d %.1lf",p->first,p->second);
                }
            }
        }
        cout<<endl;
    }
}

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