1001 A+B Format (20分) 模拟题 输出格式

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

 一刷：（c语言）

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

void print(int len ,int p[])
{
	int i = 0;
	if (len%3!=0)
	{
		for (; i < len % 3; i++)
		{
			printf("%d", p[i]);
		}
		printf(",");
	}
	int count = 0;
	for (; i < len; i++)
	{
		
		printf("%d", p[i]);
		count++;
		if (count % 3 == 0 && i < len - 1)
		{
			printf(",");
		}
	}
}

int main()
{
	int a=0, b=0;
	if(scanf("%d %d", &a, &b)==1);
	while (a < -1000000 || b>1000000)
	{
	if(scanf("%d %d", &a, &b)==1);
	}

	int isNegative = 0;
	int sum = a + b;
	if (sum<0)
	{
		isNegative = 1;
		sum = -sum;
	}

	int len = 0;
	int k = sum;
	while (k > 0)
	{
		k /= 10;
		len++;
	}

	int *p = (int *)malloc(sizeof(int)*len);

	int mask = (int)pow(10, len - 1);//分解
	if (len < 4)
	{
		if (isNegative==1)
		{
			printf("%d", -sum);
		}
		else
		{
			printf("%d", sum);
		}
		
	}
	else
	{
		for (int i = 0; i < len; i++)
		{
			p[i] = sum / mask;
			sum %= mask;
			mask /= 10;
		}

		if (isNegative == 1)
		{
			printf("-");
			print(len, p);
		}
		else
		{
			print(len, p);
		}
	}
	return 0;
}

 二刷：（c++）

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;


vector<int> v;

int main()
{
	//freopen("in.txt", "r", stdin);
	int a, b, sum;
	cin >> a >> b;
	sum = a + b;
	if (sum<0)
	{
		cout << "-";
		sum = -sum;
	}
	if (sum==0)
	{
		cout << 0 << endl;
		return 0;
	}
	while (sum!=0)
	{
		v.push_back(sum % 10);
		sum /= 10;
	}
	reverse(v.begin(), v.end());//翻转数组
	for (int i = 0; i < v.size(); i++)
	{
		cout << v[i];
		if (v.size()>3)//只有长度大于3才需要输出逗号
		{
			if ((i + 1) % 3 == v.size() % 3&&((i+1)!=v.size()))
			{
				cout << ",";
			}
		}
		
	}
	cout << endl;
	return 0;
}

 吾：java做法：（“，”逗号是分组的分隔符）见java核心（卷1）p58！c++好像不行

import java.io.Console;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		int a,b;
		Scanner in=new Scanner(System.in);
		a=in.nextInt();
		b=in.nextInt();  
		int sum=a+b;

		System.out.printf("%,d", sum);
	}
}

柳：妙啊

#include <iostream>
#include <string>

using namespace std;

int main()
{
	//freopen("in.txt", "r", stdin);
	int a, b;
	cin >> a >> b;
	string s = to_string(a + b);
	for (int i = 0; i < s.length(); i++)
	{
		cout << s[i];
		if (s[i]=='-')
		{
			continue;
		}
		if ((i+1)%3==s.length()%3&&(i!=s.length()-1))
		{
			cout << ",";
		}
	}
	return 0;
}