PAT甲级题目1-10(C++)

1001 A+B Format(20分)

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where $−10^6​​≤a,b≤10​^6$​​. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

- 

Sample Output:

-,

解题思路：

首先看到问题里面a和b的值都只在$[-10^6,10^6]$，因此用整型就足够完成加法运算，因此这里可以直接保存结果为字符串result。接下来只需要倒序遍历字符串，每3个字符就插入1个逗号，需要注意的是，最后一组不能用逗号将符号和数字分离了，比如题目里的输出：$-999,991$，就不能写成$-,999,991$。因此需要加一个判断最开头的一位是否为符号即可。

Code:

 #include<iostream>
 #include<string>
 using namespace std;

 int main(){
     int a,b;
     cin>>a>>skipws>>b;
     string result = to_string(a+b);
     int number = ;
     int length = result.length();
     string comma = ",";
     for(int i = length-;i>=;i--){
         if(number %  ==  && isdigit(result[i-])){
             result = result.insert(i,comma);
             number = ;
         }
         else{
             number++;
         }
     }
     cout<<result<<endl;
     return ;
 }

1002 A+B for Polynomials (25 分)

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 $a_{N_i}(i = 1,2,…,K)$ are the exponents and coefficients, respectively. It is given that $1 \leq K \leq 10, \ 0 \leq N_k < … < N_2 < N_1 \leq 1000.$

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:

  2.4  3.2
  1.5  0.5

Sample Output:

  1.5  2.9  3.2

解题思路：

这道题关键是要读懂题目里面说的是什么，是两个多项式相加，那就是次数不变，系数相加，根据给出的$K$的范围容易得出一开始要设定的数组长度。用数组索引作次数，数组存放的内容做系数，然后相加完以后再倒序遍历即可。

坑点：

测试点2和测试点5：这道题如果在(指数 系数)后面加空格的话，由于你不知道这个多项式的和的结果是否有常数项，如果没有常数项，那就会在Output中多了一个末尾的空格。因此要设置一个是否已经有输出的标识符，判断在输出(指数 系数)时前面是否需要加空格。

测试点6：$A+B=0$，对0要单独作额外的分类即可。

Code:

 #include<iostream>
 #include<string>
 #include<iomanip>
 using namespace std;
 double A[] = {  };
 double B[] = {  };
 int main() {
     int items;
     cin >> items >>skipws;
     int times;
     double coff;
     for (int i = ; i < items; i++) {
         cin >> times >> skipws >> coff >> skipws;
         A[times] = coff;
     }
     cin >> items>>skipws;
     for (int i = ; i < items; i++) {
         cin >> times >> skipws >> coff >> skipws;
         B[times] = coff;
     }
     int number = ;
     for (int i = ; i < ; i++) {
         A[i] += B[i];
         if (A[i] != ) {
             number++;
         }
     }
     if (number != ) {
         cout << number << " ";
         bool has_been_output = false;
         for (int i = ; i >= ; i--) {
             if (A[i] != ) {
                 if (!has_been_output) {
                     cout << i << " " << fixed << setprecision() << A[i];
                     has_been_output = true;
                 }
                 else {
                     cout << " " << i << " " << fixed << setprecision() << A[i];
                 }
             }
         }
         if (A[] != ) {
             cout << " " <<  << " " << fixed << setprecision() << A[] << endl;
         }
     }
     else {
         cout << number << endl;
     }
     return ;
 }

(未完待续)