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:

-1000000 9

Sample Output:

-999,991

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 = 1;
    int length = result.length();
    string comma = ",";
    for(int i = length-1;i>=1;i--){
        if(number % 3 == 0 && isdigit(result[i-1])){
            result = result.insert(i,comma);
            number = 1;
        }
        else{
            number++;
        }
    }
    cout<<result<<endl;
    return 0;
}

 

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 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

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[1001] = { 0 };
double B[1001] = { 0 };
int main() {
    int items;
    cin >> items >>skipws;
    int times;
    double coff;
    for (int i = 0; i < items; i++) {
        cin >> times >> skipws >> coff >> skipws;
        A[times] = coff;
    }
    cin >> items>>skipws;
    for (int i = 0; i < items; i++) {
        cin >> times >> skipws >> coff >> skipws;
        B[times] = coff;
    }
    int number = 0;
    for (int i = 0; i < 1001; i++) {
        A[i] += B[i];
        if (A[i] != 0) {
            number++;
        }
    }
    if (number != 0) {
        cout << number << " ";
        bool has_been_output = false;
        for (int i = 1001; i >= 1; i--) {
            if (A[i] != 0) {
                if (!has_been_output) {
                    cout << i << " " << fixed << setprecision(1) << A[i];
                    has_been_output = true;
                }
                else {
                    cout << " " << i << " " << fixed << setprecision(1) << A[i];
                }
            }
        }
        if (A[0] != 0) {
            cout << " " << 0 << " " << fixed << setprecision(1) << A[0] << endl;
        }
    }
    else {
        cout << number << endl;
    }
    return 0;
}

 

 (未完待续)