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 原题链接：

https://www.patest.cn/contests/pat-a-practise/1002

思路：

map映射，同时记录下表个数，同时相加相同的下标

最后整理，输出

坑点：

多项式系数为0的时候要省去。

教训：

看题不细

忽略了系数为0的情况

CODE：

#include<iostream> #include<cstring> #include<string> #include<map> #include<vector> #include<algorithm> #include<cmath> #include<cstdio> using namespace std; //计算后多项式系数为0的情况。。。 //多项式。。 map<int,double> sl; int _index[21]; bool k[1001]; bool cmp(int a,int b) { return a>b; } int main() { memset(k,0,sizeof(k)); int num=0; for (int i=0;i<2;i++) { int n; cin>>n; for (int j=0;j<n;j++) { int a; double b; cin>>a>>b; if (k[a]!=0) { sl[a]+=b; } else { _index[num]=a; num++; sl[a]=b; k[a]=1; } } } sort(_index,_index+num,cmp); int nuum=num; for (int i=0;i<num;i++) if (fabs(sl[_index[i]]-0)<0.05) nuum--; cout<<nuum; for (int i=0;i<num;i++) { if (fabs(sl[_index[i]]-0)>=0.05) printf(" %d %.1lf",_index[i],sl[_index[i]]); } return 0; }