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1002. 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 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (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


比较的时候要有fabs()    double的绝对值  #include<math.h>



评测结果

时间

结果

得分

题目

语言

用时(ms)

内存(kB)

用户

7月17日

20:51

答案正确

​​25​​

​​1002​​

​​C++ (g++ 4.7.2)​​

1

308

​​datrilla​​

测试点

测试点

结果

用时(ms)

内存(kB)

得分/满分

0

答案正确

1

308

13/13

1

答案正确

1

308

2/2

2

答案正确

1

180

2/2

3

答案正确

1

308

2/2

4

答案正确

1

308

2/2

5

答案正确

1

308

2/2

6

答案正确

1

308

2/2

#include<iostream>
#include<iomanip>
#include<math.h>
using namespace std;
#define esp 0.04
int main()
{
double index[1001]={0};
int n,a,count;
double e;
cin>>n;
while(n--)
{
cin>>a>>e;
index[a]=e;
}
cin>>n;
while(n--)
{
cin>>a>>e;
index[a]+=e;
}
n=1001;
count=0;
while(n--)
{
if(fabs(index[n])>esp)count++;
}
cout<<count;
n=1001;
while(n--)
{
if(fabs(index[n])>esp){
cout<<" "<<n<<" ";
cout<<setiosflags(ios::fixed)<<setprecision(1)<<index[n];}
}
cout<<endl;
return 0;
}


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