PAT 1002. A+B for Polynomials (25) 简单模拟

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

---

思路：

简单题目。注意格式。如果某一项系数为0，不输出。最后一种情况，没有存在项，输出"0\n"

源代码：

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 int main() {
     int na,nb;
     scanf("%d",&na);
     int flag[]={};
     double res[]={};
     int index;double data;
     for(int i=;i<na;++i) {
         scanf("%d %lf",&index,&data);
         flag[index]=;
         res[index]+=data;
     }
     scanf("%d",&nb);
     for(int i=;i<nb;++i) {
         scanf("%d %lf",&index,&data);
         flag[index]=;
         res[index]+=data;
     }
     int cnt=;
     for(int i=;i<=;++i) {
         if(flag[i]&&fabs(res[i])>1e-)
             cnt++;
     }
     if(cnt==) {
         printf("");
     } else {
         printf("%d ",cnt);
     }
     for(int i=;i>=;--i) {
         if(flag[i]&&fabs(res[i])>1e-) {
             printf("%d %.1lf",i,res[i]);
             cnt--;
             if(cnt==) {
                 break;
             } else {
                 printf(" ");
             }
         }
     }
     printf("\n");
     return ;
 }