【PAT甲级】1002 A+B for Polynomials (25分)
【PAT甲级】1002 A+B for Polynomials (25分)
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 0.5Sample Output:
3 2 1.5 1 2.9 0 3.2三个个坑
一个是 系数为0 不能算进去
一个是 输出格式又要去保留一位小数点Please be accurate to 1 decimal place.
一个是 不能==0 判断 double （其实可以 哈哈）但是 我用 abs（num-0）<0.005 不对 还是精度太小 if( abs(num[i]) >= 1e-15 )
所以要判断一个单精度浮点数：则是if( abs(f) <= 1e-6)； 要判断一个双精度浮点数：则是if( abs(f) <= 1e-15 )；
自己的答案
#include<iostream>
#include<set>
using namespace std;
double num[10005];
int main(){
int n;
set<int>s;
cin>>n;
for(int i=0;i<n;i++){
int a ;
double b;
cin>>a>>b;
num[a]+=b;
}
cin>>n;
for(int i=0;i<n;i++){
int a ;
double b;
cin>>a>>b;
num[a]+=b;
}
int count=0;
for(int i=0;i<10005;i++){
if(num[i]!=0){
count++;
}
}
cout<<count;
for(int i=10000;i>=0;i--){
if(num[i]!=0){
printf(" %d %0.1f", i, num[i]);
//cout<<" "<<i<<" "<<num[i];
}
}
return 0;
} 网搜答案
#include<stdio.h>
const int max_n = 1111;
double a[max_n] = {};
int main(){
int k, n, count = 0;
double e;
scanf("%d", &k);
for(int i = 0; i < k; i ++){
scanf("%d%lf", &n, &e);
a[n] = e;
}
scanf("%d", &k);//这里k只是为了存储，所以这儿k是可以重复存放值。
for(int i = 0; i < k; i ++){
scanf("%d%lf", &n, &e);
a[n] += e;//这儿其实就是有相同的就相加，没有相同的就不想加
}
for(int i = 0; i < max_n; i ++){
if(a[i] != 0){
count ++;
}
}
printf("%d", count);
for(int i = max_n - 1; i >=0; i --){
if(a[i] != 0) printf(" %d %.1f", i, a[i]);
}
return 0;
}