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PTA1002
阅读量:213 次
发布时间:2019-02-28

本文共 1423 字,大约阅读时间需要 4 分钟。

PTA1002 A+B for Polynomials

问题描述:

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​N1 N2 a​N​2 … N​K aN​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<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
Sample Output:
3 2 1.5 1 2.9 0 3.2

题目分析:

一开始读题没有读懂,看了别人的分析才看出来是指数排列。

输入时按照将相同底数的指数相加,遍历统计非0组的个数,再次遍历按格式输出即可,注意特殊的0。
代码如下:

#include
using namespace std;const int N=1e3+1;double a[N];int main(){ int n,x; double y; cin>>n; for (int i = 0; i < n; i++) { cin>>x>>y; a[x]+=y; } cin>>n; for (int i = 0; i < n; i++) { cin>>x>>y; a[x]+=y; } int len=0; for (int i = 0; i <= 1000; i++) { if (a[i]!=0.0) { len++; } } if(len) { int i; cout<
<<' '; for ( i = 1000; i >= 0; i--) { if (a[i]!=0.0) { printf("%d %.1f",i,a[i]); len--; if(len) cout<<' '; } } } else { cout<<"0"; } return 0;}

总结

这道题主要的难点还是读题,看懂题后很好做,通过率低的原因还是对于英文题面的理解。

转载地址:http://gfbs.baihongyu.com/

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