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 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 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
以上是题目
以下是代码
#include
<iostream>
#include
<map>
#include
<cstdio>
using
namespace
std;
int
main(){
map<
int,
float> sum;
int size_a,size_b;
cin >> size_a;
int exponent;
float coefficient;
for(
int i=
0; i<size_a; i++){
cin >> exponent >> coefficient;
sum[exponent] = sum[exponent]+coefficient;
}
getchar();
cin >> size_b;
int cnt = size_a + size_b;
for(
int i=
0; i<size_b; i++){
cin >> exponent >> coefficient;
if (sum[exponent]!=
0)
cnt--;
sum[exponent] = sum[exponent]+coefficient;
if (sum[exponent]==
0)
cnt--;
}
cout << cnt;
for(
int i=
1000; i >=
0; i--){
if(sum[i] !=
0)
printf(
" %d %.1f",i,sum[i]);
}
cout << endl;
return
0;
}
小结：这次用了一个map容器主要是想重新温习一下c++的相关内容，完全可以用一个数组来解决，相对便捷。本题主要注意一个小坑，多项式指数相同相加如果和为零，结果的多项式少两项，不是一项！
