1002 A+B for Polynomials (25分)
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
多项式相加问题。
范围用int和double足够。
申请一个容量为1001的int数组,用空间换时间,逆序输出结果。
使用里面的fixed和setprecision(n)进行格式限定。
踩坑
输出count的时候,cout<<count<<" ";
没有考虑到
1 1 N
1 1 -N
这样的情况,此时正好抵消,count=0,只输出一个0,后面没有空格。
新学的东西
cout<<i<<" "<<fixed<<setprecision(1)<<a1[i];
对a1[i],保留小数点右边一位输出。
对应的c++头文件是iomanip。
cout.precision(n),setprecision(n)可控制输出流显示浮点数的数字个数。C++默认的流输出数值有效位是6。
代码
#include<iostream>
#include<iomanip>
/*iomanip里面有setprecision(n)*/
using namespace std;
int main(){
int k1,k2;
double a1[1001]={0};
double a2[1001]={0};
cin>>k1;
int tmp=0;
for(int i=0;i<k1;i++){
cin>>tmp;
cin>>a1[tmp];
}
cin>>k2;
for(int i=0;i<k2;i++){
cin>>tmp;
cin>>a2[tmp];
}
int count=0;
int end=0;
for(int i=1000;i>=0;i--){
a1[i]=a1[i]+a2[i];
if(a1[i]!=0){
count++;
end=i;
}
}
cout<<count;
if(count!=0){
cout<<" ";
}
for(int i=1000;i>=0;i--){
if(a1[i]!=0){
cout<<i<<" "<<fixed<<setprecision(1)<<a1[i];
if(i!=end){
cout<<" ";
}
}
}
}