PAT甲级练习题- A+B for Polynomials (25)

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本文介绍了一种多项式加法的实现方法，通过输入两个多项式的系数和指数，计算并输出它们的和。代码采用C++编写，实现了多项式相加的功能，并确保输出结果保留一位小数。

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1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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

/**
题意：多项式a+b求解，结果保留一位小数，输出时按指数递减输出,
*/

#include<bits/stdc++.h>

using namespace std;
typedef pair<int,int> P;
const int maxn = 1000+10;
const int INF = 0x3f3f3f3f;

double a[maxn],b[maxn];

int main(){
    int k,t;
    
    scanf("%d",&k);
    for(int i = 0;i<k;++i){
        scanf("%d",&t);
        scanf("%lf",&a[t]);
    }
    scanf("%d",&k);
    for(int i = 0;i<k;++i){
        scanf("%d",&t);
        scanf("%lf",&b[t]);
    }

    vector<int> ans;
    for(int i = maxn;i>=0;--i){
        b[i] = a[i]+b[i];
        if(b[i]==0) continue;
        ans.push_back(i);
    }
    cout<<ans.size();
    for(int i = 0;i<ans.size();++i){
        printf(" %d %.1f",ans[i],b[ans[i]]);
    }

    return 0;
}