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
看范围发现，int型是完全ok的，所以用int就行了，相加后再转为string，来分析什么时候加逗号，将string与3取余，将其分成3的段，那么最前面剩下的就是不够的，每当i=余数时，先输出，号，之后就是每过3个输出一个逗号，相当于于3取余后仍是之前的数。
#include<bits/stdc++.h> using namespace std; int main() { int a,b; cin>>a>>b; string c=to_string(a+b); for(int i=0;i<c.length();i++){ cout<<c[i]; if(c[i]=='-') continue; if(c.length()%3==(i+1)%3&&i!=c.length()-1) cout<<","; } }