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 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

这是一个多项式相加问题，此问题类似于哈希映射，这里可以用一个一维数组来解决，

指数作为下标，底数作为值，最后遍历数组，当值不为0时，输出下标和值。

代码：

#include<iostream>#include<cstdlib>using namespace std;int main(){int num1=0,num2=0;double* a;double* b;double c[1001]={0};cin>>num1;a=(double *)malloc(num1*2*sizeof(double));for(int i=0;i<2*num1;i++){cin>>a[i];if(i%2==1){c[(int)a[i-1]]+=a[i];}}cin>>num2;b=(double *)malloc(2*num2*sizeof(double));for(int i=0;i<2*num2;i++){cin>>b[i];if(i%2==1){c[(int)b[i-1]]+=b[i];}}int y=0;for(int i=1000;i>=0;i--){if(c[i]==0) continue;else{y++;}}cout<<y<<" ";int yy=0;for(int i=1000;i>=0;i--){if(c[i]==0) continue;else{yy++;if(yy!=y){cout<<i<<" ";printf("%.1lf ",c[i]);}else{cout<<i<<" ";printf("%.1lf",c[i]);} }}return 0;}