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
二、代码
#include<iostream>
#include<string.h>
#include<cstdio>
#include<iomanip>
using namespace std;
#define max 1000
double input[max + 1];
int main()
{
int k,count;
int i,expo; //指数
double coef; //系数
//第一行多项式
cin>>k;
memset(input,0,sizeof(input)); //变量初始化
for(i=0; i<k; i++)
{
cin>>expo>>coef;
input[expo] += coef;
}
//第二行多项式
cin>>k;
for(i=0; i<k; i++)
{
cin>>expo>>coef;
input[expo] += coef;
}
//统计多项式和的个数
count = 0;
for(i=1000; i>=0; i--)
{
if(input[i] != 0)
{
count++;
}
}
cout<<count;
for(i=1000; i>=0; i--)
if(input[i] != 0.0)
{
cout<<fixed<<setprecision(1);
//加fixed表示固定点方式显示,这里的精度指的是小数位;不加fixed,则表示小数的总位数
cout<<" "<<i<<" "<<input[i];
}
cout<<endl;
return 0;
}