【PAT甲級】1002 A+B for Polynomials
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
Sample Output:
3 2 1.5 1 2.9 0 3.2
個人思路
這題就是求多項式的和,兩種思路,一種是用兩個陣列存,然後最後逐個比較最後輸出,還有一種就是用一個大小為1000的double陣列儲存指數為i,係數為poly[i]的項,在輸入的時候就合併到其中。最後從大到小輸出即可。我實現的是第二種思路。
有一個小坑,就是題目要求精確到1位小數,已在題目中加粗。
程式碼實現
#include <cstdio> #include <cstring> #include <string> #include <set> #include <map> #include <vector> #include <cmath> #include <algorithm> #include <iostream> #define ll long long #define ep 1e-5 #define INF 0x7FFFFFFF const int maxn = 1005; using namespace std; int main() { double poly[maxn]; memset(poly, 0, sizeof(poly)); int k1, k2, k = 0; // 輸入第一個多項式 cin >> k1; for (int i = 0; i < k1; i ++) { int idx; cin >> idx; cin >> poly[idx]; } // 在輸入第二個多項式的同時合併 cin >> k2; for (int i = 0; i < k2; i ++) { int idx; double tmp; cin >> idx >> tmp; poly[idx] += tmp; } // 統計多項式和的項數 int max = 0; for (int i = 0; i < maxn; i ++) { if (poly[i] != 0) { k ++; max = i; } } // 輸出 cout << k; for (int i = max; i >= 0; i --) { if (poly[i] != 0) { printf(" %d %.1lf",i, poly[i]); } } return 0; }
總結
學習不息,繼續加油