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 "

源代码：

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main() {
    int na,nb;
    scanf("%d",&na);
    int flag[1001]={0};
    double res[1001]={0};
    int index;double data;
    for(int i=0;i<na;++i) {
        scanf("%d %lf",&index,&data);
        flag[index]=1;
        res[index]+=data;
    }
    scanf("%d",&nb);
    for(int i=0;i<nb;++i) {
        scanf("%d %lf",&index,&data);
        flag[index]=1;
        res[index]+=data;
    }
    int cnt=0;
    for(int i=0;i<=1000;++i) {
        if(flag[i]&&fabs(res[i])>1e-6)
            cnt++;
    }
    if(cnt==0) {
        printf("0");
    } else {
        printf("%d ",cnt);
    }
    for(int i=1000;i>=0;--i) {
        if(flag[i]&&fabs(res[i])>1e-6) {
            printf("%d %.1lf",i,res[i]);
            cnt--;
            if(cnt==0) {
                break;
            } else {
                printf(" ");
            }
        }
    }
    printf("
");
    return 0;
}