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[PAT] A1002 A+B for Polynomials (23分!)

A1002 A+B for Polynomials Input SpecificationOutput SpecificationSample InputSample OutputCode 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

PAT-A1002 A+B for Polynomials (25分)

原题 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 a​N​1​​ N​2 aN

A1002. A+B for Polynomials

题目描述 　　This time, you are supposed to find A+B where A and B are two polynomials 输入格式 　　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

PAT A1002 A+B for Polynomials(25)

AC代码 转载自https://www.cnblogs.com/zjutJY/p/9413766.html #include <stdio.h> #include<string.h> //输入：两行数据，每行表示一个多项式：第一个数字表示非零项的数目， //后面每两个数表示一项，分别表示幂次和系数。 //输出：两个多项式的和，格式与输入一样 int main(){ const in

A1002 字符串长度

在右侧我们给出了一个已经基本完成的程序，读入了一个字符串，调用了一个叫str_len的函数来计算这个字符串的长度，并输出。 聪明的你应该已经发现了，这个叫str_len的函数并没有完成，在不修改函数原型的情况下，请完成str_len函数，实现我们上述的功能吧。 样例输入复制 abcdefg 样例输出