【PAT Advanced Level】1002. A+B for Polynomials (25)

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思路:先将两个多项式存入到两个数组中,再利用归并排序的方法合并两个多项式。

注意:遇到相加后系数为0的项要舍弃,项的数目也不能+1。

改进:写到一半发现项的指数给定了范围[0...1000],那么可以使用间址技巧,将项的指数看成数组的下标,这样扫描一遍就可以得到结果。

#include <iostream>#include <fstream>#include <vector>#include <iomanip>using namespace std;struct item{item(){}item(int e, double c){this->exp = e;this->coef = c;}int exp;double coef;};void merge(item *items1, item *items2, int a, int b);int main(){//fstream cin("a.txt");int a,b;cin>>a;int tmp = a;item *items1 = new item[a];while (tmp--){int exp;double coef;cin>>exp>>coef;item it(exp, coef);items1[a - tmp - 1] = it;}cin>>b;tmp = b;item *items2 = new item[b];while (tmp--){int exp;double coef;cin>>exp>>coef;item it(exp, coef);items2[b - tmp - 1] = it;}merge(items1, items2, a, b);return 0;}void merge(item *items1, item *items2, int a, int b){vector<item> v;int i = 0,j = 0;for(;i < a && j < b;){if(items1[i].exp > items2[j].exp){if(items1[i].coef != 0)v.push_back(items1[i]);i++;}else if(items1[i].exp < items2[j].exp){if(items2[j].coef != 0)v.push_back(items2[j]);j++;}else{item it(items1[i].exp, items1[i].coef + items2[j].coef);if (it.coef != 0)v.push_back(it);i++;j++;}}if(i == a){while (j < b){if(items2[j].coef != 0)v.push_back(items2[j]);j++;}}else{while (i < a){if(items1[i].coef != 0)v.push_back(items1[i]);i++;}}cout<<v.size();for (int i = 0; i < v.size(); i++){cout<<" "<<v[i].exp<<" "<<fixed<<setprecision(1)<<v[i].coef;}cout<<endl;}


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