PAT Advanced 1002 A+B for Polynomials (25 分) c++语言实现(g++)

2021-05-29 10:03

阅读：761

标签：cti key code 测试 man 注意去除 push 重复

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

多项式求和

输入为两行一个多项式的项数k 然后给出k组 e次幂和 a系数

输出一行同样　　多项式项数k k组 e次幂和 a系数

本身不难,但是要注意当同幂次系数相加和为0 时,需要删除这一项

这里用的是mapint 是幂,double是值 , 直接删除 key-value 键值对, 用map.erase(iterator);参数为指向key的迭代器

测试点 3 4 5 6 当最终结果没有数据的时候, 只输出项数 0 即可, 不需要输出多项式

附一个来自其他同学的测试用例https://blog.csdn.net/han_hhh/article/details/106338514

input
0
0
output
0
 
input
1 1000 1.5
0
output
1 1000 1.5
 
input
1 1000 1.5
1 1000 -1.5
output
0
 
input
2 1000 1.5 500 1.5
1 1000 -1.5
output
1 500 1.5

AC后去除了重复多余代码的版本

#include 
#include 
#include 
#include using namespace std;
void getPolynomial(mapint,double> &a){
    int k,ni;
    double ai;
    cin >> k;
    mapint,double>::iterator it;
    for(int i=0;i){
        cin >> ni >> ai;
        a[ni]+=ai;
        if(a[ni]==0)
        {
            it=a.find(ni);
            a.erase(it);
        }
    }
}
int main(){
    mapint,double>::iterator it;
    mapint,double> polynomial;
    vectorint> ni;
    vectordouble> ai;
    getPolynomial(polynomial);
    getPolynomial(polynomial);
    for(it=polynomial.begin();it!=polynomial.end();it++){
        ni.push_back(it->first);
        ai.push_back(it->second);
    }
    cout  ai.size();
    cout.setf(ios::fixed);
    for(int i=ni.size()-1;i>=0;i--){
        cout " "" "1) ai[i];
    }
    return 0;
}

第一次提交

//多项式求和
#include 
#include 
#include 
#include using namespace std;
void getPolynomial(mapint,double> &poly){
    int k,ni;
    double ai;
    cin >> k;
    for(int i=0;i){
        cin >> ni >> ai;
        poly[ni]=ai;
    }
}
void aplusb(mapint,double> &a,mapint,double>&b){
    mapint,double>::iterator it;
    mapint,double>::iterator itb;
    for(it=a.begin();it!=a.end();it++){
        itb=b.find(it->first);
        b[it->first]+=it->second;
        if(b[it->first]==0)
        {
            b.erase(itb);
        }
    }
}
int main(){
    mapint,double>::iterator it;
    mapint,double> polynomial1;
    mapint,double> polynomial2;
    mapint,double> result;
    vectorint> ni;
    vectordouble> ai;
    getPolynomial(polynomial1);
    getPolynomial(polynomial2);
    aplusb(polynomial1, result);
    aplusb(polynomial2, result);
    for(it=result.begin();it!=result.end();it++){
        ni.push_back(it->first);
        ai.push_back(it->second);
    }
    cout  ai.size();
    cout.setf(ios::fixed);
    for(int i=ni.size()-1;i>=0;i--){
        cout " "" "1) ai[i];
    }
    return 0;
}

PAT Advanced 1002 A+B for Polynomials (25 分) c++语言实现(g++)

标签：cti key code 测试 man 注意去除 push 重复

原文地址：https://www.cnblogs.com/ichiha/p/14771082.html

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PAT Advanced 1002 A+B for Polynomials (25 分) c++语言实现(g++)

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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