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

   题目只是要求按照英文格式读出数字，又由于数字范围限制在-10^6~10^6之间,因此可以采用两种做法:

#include
#include
#include
#include

using namespace std;
//思路：先转换再反转添加,

int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    int c=a+b;
    string s=to_string(c);
    if(c<0){
        printf("-");
        s.erase(s.begin());
    }
    reverse(s.begin(),s.end());
    string ans;
    for(int i=0;i<(int)s.length();i++){
        if(i%3==0&&i!=0){
            ans+=",";
            ans+=s[i];
        }
        else ans+=s[i];
    }
    reverse(ans.begin(),ans.end());
    cout<     return 0;
}
 

第二种,直接讨论输出","位置:

#include
#include
#include
#include

using namespace std;

int main()
{
    int a,b,c;
    scanf("%d%d",&a,&b);
    c=a+b;
    if(c<0){        //调整负号
        printf("-");
        c=-c;}
    if(c/1000000!=0)   printf("%d,%03d,%03d",c/1000000,c/1000%1000,c%1000);
    else if(c/1000!=0)  printf("%d,%03d",c/1000,c%1000);
    else   printf("%d",c);
    return 0;
}