pta甲级1002 A+B for Polynomials（AC）

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 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 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

结尾无空行

Sample Output:

3 2 1.5 1 2.9 0 3.2

结尾无空行

思路：
不能说很简单，只能说完全不要动脑子。

#include<bits/stdc++.h>
using namespace std;
double a[1001][3],b[1001];
int main()
{
	int k,count=0;
	for(int i=1;i<=2;i++){
		cin>>k;
		for(int j=1;j<=k;j++){
			int m;
			cin>>m;
			cin>>a[m][i];
		}
	}
	for(int i=0;i<=1000;i++){
		b[i]=a[i][1]+a[i][2];
		if(b[i]!=0) count++;
	}
	printf("%d",count);
	for(int i=1000;i>=0;i--){
		if(b[i]!=0) printf(" %d %.1lf",i,b[i]);
	}
}

 搞定！！

标签：case,AC,int,pta,each,input,line,where,1002
来源： https://blog.csdn.net/worldinme/article/details/120896719