C++甲级1001 A+B Format （20 分）（sprintf,memset,reverse）

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
 
int main()
{
    int a,b;
    while(scanf("%d%d",&a,&b)==2)
    {
        int c=a+b;
        char ans[9];//-1000000
        sprintf(ans,"%d",c);////把整数c 打印成一个字符串保存在ans中。头文件：cstdio 
        char res[10];//-1,000,000
        memset(res,0,sizeof(res));//将res中当前位置后面的sizeof(res)个字节 用 0 替换并返回 res ,相当于初试化res数组.头文件：cstring 
        int len=strlen(ans);
        int j=0;
        int cnt=0;
        for(int i=len-1;i>=0;i--)//因为输出是从后往前每三个数一个逗号，所以要借助数组res,逆序保存结果 ，再反转数组最后输出 
        {
            res[j++]=ans[i];
            cnt++;
            if(ans[i-1]>='0'&&ans[i-1]<='9' && cnt%3==0)
            {
                res[j++]=',';
            }
        }
        reverse(res,res+j);//reverse(res[0],res[j]) 可以将数组指针在[0,j)之间的元素范围内的元素进行反转 头文件：algorithm 
        printf("%s\n",res);
 
 
    }
    return 0;
}

F2:借助栈先进后出的特点 

#include<cstdio>
#include<stack>
using namespace std;
int main()
{
	int a,b,st;
	scanf("%d%d",&a,&b);
	st = a + b;
	int i = 0;
	if(st<1000 && st>-1000)//若为三位数则直接输出 
	{
		printf("%d\n",st);
		return 0;
	}

	if(st < 0)
	{
		printf("-");
		st=-st;
	}
	stack<int> sta;//建立一个栈，栈为先进后出 ，头文件<stack> 
	while(st>0)
	{
		i++;
		sta.push(st%10);//个位数依次进栈 
		st=st/10;
	}
	while(!sta.empty())
	{
		i--;
		printf("%d",sta.top());//输出栈顶元素 
		if(i!=0&&i%3==0)
			printf(",");
		sta.pop();//将栈顶元素删除 
 
	}
	printf("\n");
	return 0;
}