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这道题主要是让你初始输出，一般比较长的数字我们转化为字符串处理 如果是负数那么“-”照样输出 因为是从后往前数每3位要打一个“，”，那么就找第一个不满3位的且不是最后一位的，在它后面打“，”， 因为i是序号，真正的位数是i+1位，如果字符串为4，那么多出的就是4%3=1，就是一位，那么只要找出前1位，也就是(i+1)%3 == 4%3的那一位就行了
#include<cstdio> #include<iostream> using namespace std; int main(){ int a, b; scanf("%d%d",&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; }