PAT_A 1002 A+B for Polynomials

PAT_A 1002 A+B for Polynomials

分析

本题是多项式相加，按指数非递增顺序输入两多项式，对应次数相加即可，需要注意的是一多项式中可能会有多项次数相同的项，不能简单的赋值；还需要注意特殊情况，如多项式相加后消去的情况。

这里找到了某大佬的测试用例如下，可以辩证的参考借鉴。

某大佬的测试用例如下

我的全部测试

test	result
2 1 2.4 0 3.2 2 2 1.5 1 0.5	3 2 1.5 1 2.9 0 3.2
3 2 2.5 1 2.4 0 3.2 2 2 1.5 1 0.5	3 2 4.0 1 2.9 0 3.2
1 0 3.2 1 0 -3.2	0
1 0 3.2 1 0 7.89	1 0 6.1
3 2 2.5 1 2.4 0 3.2 2 2 -2.5 1 0.5	2 1 2.9 0 3.2
3 2 2.5 2 2.4 0 3.2 2 2 -2.5 1 0.5	3 2 2.4 1 0.5 0 3.2
10 2 2.5 2 2.4 0 3.2 2 2.5 2 2.4 0 3.2 2 2.5 2 2.4 0 3.2 0 4 10 2 -2.5 1 0.5 2 -2.5 1 0.5 2 -2.5 1 0.5 2.5 2 2.4 0 3.2 0 4	3 2 10.1 1 1.5 0 20.8

题目的描述

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:

\(K\ N_1\ a_{N_1}\ N_2\ a_{N_2}\ ...\ N_K\ a_{N_K}\)

where K is the number of nonzero terms in the polynomial, \(N_i\ and\ a_{N_i}\) (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤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

AC的代码

#include<bits/stdc++.h>

using namespace std;

int main(){
    int K,i=0,c=0;
    array<float, 1001> A{0},B{0};
    cin>>K;
    while(K){
        K--;
        int t=0;
        float t1;
        cin>>t>>t1;
        A[t]+=t1;
    }
    cin>>K;
    while(K){
        K--;
        int t=0;
        float t1;
        cin>>t>>t1;
        B[t]+=t1;
    }
    for(int i=0;i<1001;i++){
        A[i]+=B[i];
        if(A[i]<=-0.1||A[i]>=0.1)c+=1;
    }
    cout<<c;
    for(int i=1000;i>-1;i--){
        if(A[i]<=-0.1||A[i]>=0.1){
            printf(" %d %.1f",i,A[i]);
        }
    }
    cout<<endl;
    
    return 0;
}