【PAT】1002. A+B for Polynomials (25)

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1002

分析：给出俩个多项式的指数和系数，没有给出基数，只要将指数相同的多项式进行系数相加即可。输出时按照指数从高到低的顺序输出。

        

题目描述：

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

参考代码：

#include<iostream>
#include<string.h>
using namespace std;
#define max 1000
double input[max + 1];

int main()
{
    int k,count;
    int i,expo;
    double coef;
    scanf("%d",&k);
    memset(input,0,sizeof(input));
    for(i=0; i<k; i++)
    {
        scanf("%d %lf",&expo,&coef);
        input[expo] += coef;
    }
    cin>>k;
    for(i=0; i<k; i++)
    {
        scanf("%d %lf",&expo,&coef);
        input[expo] += coef;
    }

    count = 0;
    for(i=1000; i>=0; i--)
        if(input[i] != 0) count++;

    printf("%d",count);
    for(i=1000; i>=0; i--)
        if(input[i] != 0.0)
            printf(" %d %.1lf",i,input[i]);
    cout<<endl;
    return 0;
}

以上代码，在最后输出的时候，还是要从1000遍历到0。这其中可能很多都是空的，浪费了遍历时间。所以把输出进行了修改，写了一份新的代码如下：

#include<iostream>
#include<stack>
#include<string.h>
using namespace std;
#define max 1001
double arrayA[max];
struct Node{
Node(int e, double c){
exp = e;
coe = c;
}
int exp;
double coe;
};

int main(){
int cntA, i, exp;
double coe;
memset(arrayA,0,sizeof(arrayA));
cin>>cntA;
for(i=0; i<cntA; i++){
cin>>exp>>coe;
arrayA[exp] = coe;
}
cin>>cntA;
for(i=0; i<cntA; i++){
cin>>exp>>coe;
arrayA[exp] += coe;
}
int cnt = 0;
stack<Node> sta;
for(i=0; i<max; i++){
if(arrayA[i] != 0){
Node n(i,arrayA[i]);
sta.push(n);
}
}
cout<<sta.size();
if(sta.size() != 0){
while(!sta.empty()){
printf(" %d %.1lf", sta.top().exp, sta.top().coe);
sta.pop();
}
}
cout<<endl;
return 0;
}