PAT-1001 A+B Format 解答(C/C++/Java/python)

1.Description：

 

2.Example：

Input:
-100000 9
Output:
-999,991

3.Solutions:

C Version：

//
// Created by SheepCore on 2020-02-23.
//

#include <stdio.h>
#include <string.h>

int main() {
    int a, b; 
    char str[10];

    scanf("%d %d", &a, &b);
    sprintf(str, "%d", a + b);
    int len = strlen(str);
    for (int i = 0; i < len; ++i) {
        printf("%c", str[i]);
        if(str[i] == '-')
            continue;
        if((i + 1) % 3 == len % 3 && i != len -1)
            printf(",");
    }
} 

Result：

C ++ Version：

#include <iostream>
using namespace std;

int main() {
    int a, b;

    cin>> a>> b;
    string s= to_string(a+ b);
    int len= s.length();
    for (int i= 0; i< len; ++i) {
        cout<< s[i];
        if (s[i] == '-')
            continue;
        if ((i + 1) % 3 == len % 3 && i != len - 1)
            cout<< ",";
    }
}

Result：

Java Version：

package pat_online_test;

import java.util.Scanner;

/**
 * @author sheepcore
 */
public class P1001_add_format {
    public static void main(String[] args) {
        int a, b, sum;
        String s;
        Scanner scanner = new Scanner(System.in);

        a = scanner.nextInt();
        b = scanner.nextInt();
        sum = a + b;
        boolean sign = sum >= 0;
        s = sum + "";
        int len = s.length();

        for (int i = 0; i < len; i++) {
            System.out.print(s.charAt(i));
            if(s.charAt(i) == '-'){
                continue;
            }
            if((i+1) % 3 == len % 3 && i != len -1){
                System.out.print(',');
            }
        }
    }
}

Result：

 Python Version：

"""
 A+B Format
 Created by SheepCore at 2020-3-24
"""

line = input().split()
a, b = eval(line[0]), eval(line[1])
sum = a + b

sign = sum >= 0
str = str(sum)
length = len(str)

i = 0
while i < len(str):
    print(str[i], end='')
    if str[i] == '-':
        i += 1
        continue
    if (i+1) % 3 == length % 3 and i != length - 1:
        print(',', end='')
    i += 1

Result：

4.Summary:

字符打印时，注意下标的转换。

之前第一次做的时候分正负号讨论过，所以代码不算简洁，然后参考有些大神的，写了一个较为简洁的！继续加油！

笨鸟先飞，水滴石穿！