【PAT甲】1002 A+B for Polynomials (25分)（多项式相加）

problem

1002 A+B for Polynomials (25分) 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

将两个多项式相加，输出一个多项式

solution

题目范围只有1000，所以最多a0到a1000。多项式系数可以为小数，所以开个double a[1000]模拟即可。一些优化：可以mn和mx记录最小值最大值就不用遍历1000，可以输入时加个特判if(a[i]!=0)记录有几个就不用统计1000，不过数据小懒得优化了。注意输出格式要求末位不能有空格（包括最后一个点，答案为0时末位也不能有空格） #include<iostream> #include<cstdio> using namespace std; double a[1010]; int main(){ int k1, k2; cin>>k1; for(int i = 1; i <= k1; i++){ int x; double y; cin>>x>>y; a[x] += y; } cin>>k2; for(int i = 1; i <= k2; i++){ int x; double y; cin>>x>>y; a[x] += y; } int cnt = 0; for(int i = 1000; i >= 0; i--){ if(a[i] != 0)cnt++; } if(cnt!=0)cout<<cnt<<' '; else cout<<cnt; int tmp = 0; for(int i = 1000; i >= 0; i--){ if(a[i] != 0){ tmp++; if(tmp == cnt){ printf("%d %.1lf",i,a[i]); break; } printf("%d %.1lf ",i,a[i]); } } return 0; }