PAT1002:1002. A+B for Polynomials 解题报告

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 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

代码：

#include<cstdio>#include<queue>using namespace std;struct Node{  int ex;  float co;  Node(int e, float c):ex(e),co(c){}};bool isZero(float co){  if(co < 0) co = -co;  if(co < 1e-5) return true;  return false;}queue<Node> poly1;queue<Node> poly2;int main(){  int n,ex;  float co;  scanf("%d",&n);  while(n--){    scanf("%d%f",&ex,&co);    poly1.push(Node(ex,co));  }  scanf("%d",&n);  while(n--){    scanf("%d%f",&ex,&co);    poly2.push(Node(ex,co));  }  queue<Node> sum;  while(!poly1.empty() && !poly2.empty()){    Node p1 = poly1.front(), p2 = poly2.front();    if(p1.ex < p2.ex){      if(!isZero(p2.co)) sum.push(p2);      poly2.pop();    }else if(p1.ex == p2.ex){      float co = p1.co + p2.co;      if(!isZero(co)) sum.push(Node(p1.ex, co));      poly1.pop(); poly2.pop();    }else{      if(!isZero(co)) sum.push(p1);      poly1.pop();    }  }  while(!poly1.empty()){    sum.push(poly1.front());    poly1.pop();  }  while(!poly2.empty()){    sum.push(poly2.front());    poly2.pop();  }  int len = sum.size();    printf("%d",len);  while(!sum.empty()){    Node e = sum.front();    printf(" %d %.1f",e.ex,e.co);    sum.pop();  }  printf("\n");  return 0;}