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>
using namespace std;
int main(){
int x, y;
scanf("%d %d", &x, &y);
int sum = x + y;
if(sum == 0){
printf("0");
return 0;
}
if(sum < 0){
printf("-");
sum = sum * (-1);
}
const int LEN = 8 + 2 + 1;
char a[LEN];
int i = 1;
int j = 1;
while(sum > 0){
a[i++] = sum % 10 + '0';
sum = sum / 10;
if(j % 3 == 0 && sum > 0){
a[i++] = ',';
}
++j;
}
for(int k = i-1; k > 0; k--){
printf("%c", a[k]);
}
return 0;
}
377

被折叠的 条评论
为什么被折叠?



