PAT(Advanced)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:

$K$ N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ … N​K​​ a​N​K​​​​

where $K$ is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ ($i=1,2,\cdots ,K$) are the exponents and coefficients, respectively. It is given that $1\le K\le 10$，0≤N​K​​<⋯<N​2​​<N​1​​≤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

题目解读：输入A、B两个多项式数列，求这两个多项式的和以及有效项个数，保留一位小数。

解题思路： 因为最大指数不是很大，所以直接用数组储存这两个多项式，不需要分别放两个，放在一个里面加就可以了，数组最好用fill()初始化一下。速度还是有点慢，卡在了输出格式上，题目的保留一位小数用%.1f输出就可以了，我用了%g输出，区别就是%g保留的是有效数字。还在最后一个格式卡了一下，如果是答案是0的话就没有后面一个空格，所以空格应该是每输出一组多项式之前带的，改了一下就通过了。

AC代码

#include <iostream>
#include <vector>
using namespace std;
struct node{
    int a;
    float b;
    node(int n1,float n2){
        a = n1;
        b = n2;
    }
};
vector<node> result;
float ans[1003];
int main(int argc, const char * argv[]) {
    int n;
    fill(ans,ans+1003,0);
    for( int num = 0 ; num < 2 ; num ++ ){
    scanf("%d",&n);
    for( int i = 0; i < n ; i ++ ){
        int t;
        float p;
        scanf("%d %f",&t,&p);
        ans[t] = ans[t]+p;
    }
    }
    for(int i = 1002 ; i >= 0 ; i -- ){
        if(ans[i] != 0 ){
            result.push_back(node(i,ans[i]));
        }
    }
    printf("%d",result.size());
    for( int i = 0 ; i < result.size(); i ++ ){
        printf(" %d %.1f",result[i].a,result[i].b);
    }
    return 0;
}