1009. Product of Polynomials (25)

时间限制
400 ms
内存限制
32000 kB
代码长度限制
16000 B
This time, you are supposed to find A*B where A and B are two polynomials.
Input Specification:
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 Specification:
For each test case you should output the product 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 up to 1 decimal place.
Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

=====================================
这题目没什么难度的说，就是多项式乘积嘛~
唯一要注意的事情是，结果的输出时按照指数递减的…这里用了map里面的reverse_iterator，真心是好用！
另外，也不知道系数有没有负数，所以处理的时候考虑了一下相加为0的情况…貌似没啥必要哦…
总之是很简单的东西~前提是有STL这货…
=====================================

#include 
#include 
using namespace std;
#define MapIterator map::iterator
#define MapIteratorReverse map::reverse_iterator
int main()
{
    int K;
    scanf("%d", &K);
    map polyA; //第一行
    for(int i=0; i p(Ni, Ki);
        polyA.insert(p);
    }
    scanf("%d", &K);
    map result; //结果
    for(int i=0; ifirst + Ni;  //乘方
            float Kj = it->second * Ki;//系数
            if(result.count(Nj) > 0)
            {
                //原来有结果
                float sum = Kj + result[Nj];
                if(sum == 0)
                    //这项没了...
                    result.erase(Nj);
                else
                    result[Nj] = sum;
            }
            else
            {
                result.insert(pair(Nj, Kj));
            }
        }
    }
    printf("%d", result.size());
    MapIteratorReverse it;
    for(it = result.rbegin(); it != result.rend(); it++)
    {
        printf(" %d %.1f", it->first, it->second);
    }
    return 0;
}

Leave a Reply Cancel reply