PAT (Advanced Level) Practise 1002 A+B for Polynomials

最新推荐文章于 2020-05-29 11:45:18 发布

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最新推荐文章于 2020-05-29 11:45:18 发布

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本文介绍了一种解决多项式加法问题的算法实现，输入两个多项式，输出它们的和。该算法首先读取两个多项式的系数和指数，然后通过遍历数组的方式计算并输出结果。

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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

解题思路：多项式相加

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <map>

using namespace std;

const int INF=0x3f3f3f3f;
double a[1009];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof a);
        int x;
        double y;
        for(int i=0;i<n;i++)
        {
            scanf("%d %lf",&x,&y);
            a[x]+=y;
        }
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d %lf",&x,&y);
            a[x]+=y;
        }
        int sum=0;
        for(int i=0;i<1006;i++)
            if(a[i]) sum++;
        printf("%d",sum);
        for(int i=1005;i>=0;i--)
            if(a[i]) printf(" %d %.1lf",i,a[i]);
        printf("\n");
    }
    return 0;
}