a1002. A+B for Polynomials (25)

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

题意：就是普通二项式相加，因为输入的顺序已确定，从大到小，利用two pointers 很好解决，刚开始没考虑到相加为0的情况，有几个点没过。

ac代码：

#include <iostream>#include <cstdio>using namespace std;struct Poly{float expo;double coef;}a[12],b[12],ans[12];int main(){//freopen("D:\input.txt","r",stdin);int k1,k2;cin>>k1;for(int i=0;i<k1;i++){cin>>a[i].expo>>a[i].coef;}cin>>k2;for(int i=0;i<k2;i++){cin>>b[i].expo>>b[i].coef;}int cnt=0,i=0,j=0;for(;i<k1&&j<k2;){if(a[i].expo==b[j].expo){   //相同项合并 if(a[i].coef+b[j].coef!=0){ans[cnt].expo=a[i].expo;ans[cnt].coef=a[i].coef+b[j].coef;++cnt; }++i; ++j;}else if(a[i].expo>b[j].expo){ans[cnt].expo=a[i].expo;ans[cnt].coef=a[i].coef;++cnt; ++i;}else {ans[cnt].expo=b[j].expo;ans[cnt].coef=b[j].coef;++cnt; ++j;}}while(i<k1){ans[cnt].expo=a[i].expo;ans[cnt].coef=a[i].coef;++cnt; ++i;}while(j<k2){ans[cnt].expo=b[j].expo;ans[cnt].coef=b[j].coef;++cnt; ++j;}cout<<cnt;for(int i=0;i<cnt;i++){//cout<<" "<<ans[i].expo<<" "<<ans[i].coef;printf(" %1.f %.1f",ans[i].expo,ans[i].coef);}}