程序设计入门17 数组A+B Format

1001 A+B Format (20)（20 分）

Calculate a + b and output the sum in standard format -- that is, thedigits must be separated into groups of three by commas (unless thereare less than four digits).

Input

Each input file contains one test case. Each case contains a pair ofintegers a and b where -1000000 <= a, b <= 1000000. The numbersare 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

分析：本题的难点在于间隔输出逗号。因为是从低位开始间隔三位，所以开头从哪开始输出逗号是个难题。

解决这个问题要把结果逆序存储在数组里，即a[0]存个位，a[1]存十位，a[2]存百位，数组的末位开始输出，这样从下标是否是3的倍数便可输出逗号，模拟了实际位数（下标）和位数上的数字的表示。

一，正确代码：

#include<iostream>
using namespace std;
int main(){
	int num[10];
	int a,b,sum;
	int len=0;
	scanf("%d %d",&a,&b);
	sum = a+b;
	if(sum<0){
		sum=-sum;
		printf("-");
	}
	if(sum==0){
		num[len++]=0;
	}
	while(sum){
		num[len++]=sum%10;
		sum/=10;
	}
	for(int i=len-1;i>=0;i--){
		printf("%d",num[i]);
		if(i>0&&i%3==0) printf(",");
	}
	return 0;
}