PAT1002 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 N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (,) are the exponents and coefficients, respectively. It is given that 1，0.

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可耻地没读懂题目。。这道题多项式加法也太不按（我的）常理出牌了吧（嘤嘤嘤每个样例包含两行，每行第一个数表示几组（指数，系数）那么输出也是这样。第一个数表示几组（指数，系数）这边开始我不明白起来了。。为啥是3呢= =原来这个统计的是指数的个数（跪了好吧）相同指数的系数相加这个输出还有点顺序→指数从大到小排列分析完了我们就可以写了（呜呜呜） #include<iostream> #include<iomanip> #include<stdlib.h> #include<stdio.h> using namespace std; int main() { int n, ex, cnt = 0; double co; double a[10000] = { 0 }; cin >> n; for (int i = 0; i < n; i++) { cin >> ex >> co; a[ex] += co; } cin >> n; for (int i = 0; i < n; i++) { cin >> ex >> co; a[ex] += co; } for (int i = 0; i < 1001; i++) { if (a[i] != 0) { cnt++; } } cout << cnt; for (int i = 1000; i >= 0; i--) { if (a[i] != 0) { printf(" %d %.1lf", i, a[i]); } } system("pause"); return 0; }

呜呜呜，说几个卡着的点

第一次错是因为系数是浮点型（嘤嘤嘤！数组也记得开浮点型嗷

第二个就是系数记得保留一位小数！！

上面是AC代码嘤嘤嘤溜了溜了

 

 

转载于:https://www.cnblogs.com/kazama/p/10877860.html

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