pat1002. A+B for Polynomials (25)

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

---

提交代码

#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<algorithm>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<queue>
using namespace std;
double co[1005];
#define inf 1e-8
int main()
{
    //freopen("D:\INPUT.txt","r",stdin);
    int i,k,ex;
    double num;
    int sumcount=0;
    scanf("%d",&k);
    while(k--){
        scanf("%d %lf",&ex,&num);
        co[ex]=num;
    }
    scanf("%d",&k);
    while(k--){
        scanf("%d %lf",&ex,&num);
        co[ex]+=num;
    }
    for(i=1000;i>=0;i--){
        if(fabs(co[i])<inf){
            continue;
        }
        sumcount++;
    }
    printf("%d",sumcount);
    for(i=1000;i>=0;i--){
        if(fabs(co[i])<inf){
            continue;
        }
        printf(" %d %.1lf",i,co[i]);
    }
    printf("
");
    return 0;
}