PAT(Advanced level) 1001 A+B Format(20)

最新推荐文章于 2024-09-04 16:30:41 发布

RRR_mx

最新推荐文章于 2024-09-04 16:30:41 发布

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本文介绍了一种解决特定格式输出两个整数相加结果的方法。通过两种不同的实现方式，一是使用条件语句进行数字分组，二是将整数转换为字符串后按位输出，来确保输出结果符合标准格式要求，即每三位用逗号分隔。

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个人博客题解

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

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

就是简单的数字分组，三个一组，给定了范围，直接if判断就可以

#include <iostream>
using namespace std;

int main()
{
	int a, b;
	cin >> a >> b;
	int res = a + b;  printf("%d\n", res);
	if(res < 0) { res = -res; printf("-"); }
	
	if(res <= 999 && res >= 0) printf("%d", res);
	if(res >= 1000 && res <= 999999) printf("%d,%03d", res/1000, res%1000);
	if(res >= 1000000) printf("%d,%03d,%03d", res/1000000, res%1000000/1000, res%1000);
	
	return 0;
}

还看到一种方法感觉也挺方便转为字符串看下标就可以

#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 << ","; 
    } 
    return 0; 
}