PAT甲级1002

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1002. A+B for Polynomials (25)

This time, you are supposed to find A+B where A and B are two polynomials.

计算两个多项式的和。

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.

每个输入文件包含一个测试用例。每个例子有两行，每一行分别是一个多项式的信息：K N1 a_N1 N2 a_N2 ... NK a_NK。K是这个多项式中不为0的项的个数，Ni and a_Ni (i=1, 2, ..., K)是指数和系数。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.

以跟输入相同的形式输出A+B的结果。需要注意的是在每一行结尾没有多余的空格。输出精确到小数点后一位。

Sample Input

2 1 2.4 0 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

#include<stdio.h>#include<cstring>#include<math.h>#include<iostream>#include <algorithm>#include<cmath>using namespace std;int main(){float sum[1001];for(int i=0;i<1001;i++)sum[i]=0;int n=0,num=0,k=0;float nn=0.0;for(int i=0;i<2;i++){cin>>k;for(int j=0;j<k;j++){cin>>n;cin>>nn;sum[n]+=nn;}}for(int i=0;i<1001;i++){if(sum[i]!=0)num++;}cout<<num;for(int i=1000;i>=0;i--){if(sum[i]!=0){cout<<" "<<i<<" ";printf("%.1f",sum[i]);}}system("pause");return 0;}

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