1002 A+B for Polynomials
题目描述
This time, you are supposed to find A+B where A and B are two polynomials.
Input Specification:
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 Specification:
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
由于只有两行多项式,且多项式最多只有1000项,因此只用定义两个向量,将每一项的系数存在对应的索引位置,然后两个向量相加即可
注意:
说明一下多项式的形式,对于输入
- K N1 aN1 N2 aN2 … NK aNK
这个多项式是:
- aN1XN1+aN2XN2+…+aNKXNK
在样例中,2 1 2.4 0 3.2表示这个多项式有两项,指数为1的系数为2.4,指数为0的系数为3.2,即:
- 2.4X1+3.2X0
代码
代码使用两个函数,一个是初始化数据init,另一个是对数据进行处理。
初始化的时候,将第一个多项式的系数存在coefficient[0][i]中,第二个存在coefficient[1][i]中。然后对应每一项系数相加
输出前,先记录非零项的系数,输出的时候不输出系数为0的项目
注意: 要使用double数组来储存系数
#include<stdio.h>
#define MAXK 1001
double coefficients[2][MAXK];//store two lines of exponents and coefficients,index is exponent and value is coefficient
int K;
void init();
void process();
int main()
{
init();
process();
return 0;
}
void init()
{
//initialize coefficients array
for(int i=0;i<2;i++)
for(int j=0;j<MAXK;j++)
coefficients[i][j]=0;
int temp;
for(int i=0;i<2;i++)//put the number to two int[]
{
scanf("%d",&K);
for(int j=0;j<K;j++)
{
scanf("%d",&temp);
scanf("%lf",&coefficients[i][temp]);
}
}
return;
}
void process()
{
int count=0;
for (int i=0; i<MAXK; i++)
{
coefficients[0][i]+=coefficients[1][i];
if(coefficients[0][i]!=0)//recording the count of the nonezero terms
count++;
}
printf("%d",count);
for(int i=MAXK-1;i>=0;i--)
{
if(coefficients[0][i]!=0)
printf(" %d %.1lf",i,coefficients[0][i]);
}
printf("\n");
return;
}
git仓库:A+B for Polynomials