PAT 1002 A+B for Polynomials（map模拟）

This time, you are supposed to find A+B where A and B are two polynomials（多项式）.

Input

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents（指数） and coefficients（系数）, 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

题目意思：两个多项式的相加，每一行第一个数k表示非零系数的个数，之后是每个非零系数的指数和系数。

解题思路：可以直接使用数组来模拟，这里我使用map复习一下map的用法。

https://www.cnblogs.com/wkfvawl/p/9387566.html

#include

#include

#include

#include

#include

using namespace std;

int main()

{

int T=2;

int n,k,i;

double m;

int cnt=0;

mapmp;

map::iterator it;//迭代器

map::reverse_iterator rit;//反向迭代器

while(T--)

{

scanf("%d",&k);

for(i=0; i

{

scanf("%d%lf",&n,&m);//n为指数,m为系数

mp[n]+=m;

}

}

for(it=mp.begin();it!=mp.end();it++)//系数为0的

{

if(it->second!=0)

{

cnt++;

//printf("%lf\n",it->second);

}

}

printf("%d",cnt);

//map默认从小到大，可以用逆向迭代器逆序输出

for(rit=mp.rbegin();rit!=mp.rend();rit++)//系数为0的

{

if(rit->second!=0)

{

printf(" %d %.1f",rit->first,rit->second);

}

}

printf("\n");

return 0;

}

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