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 Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample Output
3 2 1.5 1 2.9 0 3.2
解析：其实就是简单的多项式相加，将相同指数项的coefficient相加即可，注意零项需要去除。也可以使用key-value，这样可以节省内存。
Code:
#include <iostream> #include <iomanip> #include <string.h> #include <vector> #include <algorithm> #include <stdio.h> using namespace std; int main(){ //use array to store double A[1100]; memset(A,0,sizeof(A)); int k1; cin>>k1; int ex; double co; while(k1--){ cin>>ex>>co; A[ex] = co; } int k2; cin>>k2; int ex2; double co2; while(k2--){ cin>>ex2>>co2; A[ex2] += co2; } //serach for non-zero vector<int> exs; vector<double> cos; int count = 0; for(int i=1000; i>=0; i--){ if(A[i] != 0){ exs.push_back(i); cos.push_back(A[i]); count++; } } cout<<count; for(int i=0; i<exs.size(); i++){ printf(" %d %.1lf",exs[i],cos[i]); } return 0; }
原文：http://www.cnblogs.com/RookieCoder/p/5023680.html