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 −106≤a,b≤106. 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
Submit:
#include <stdio.h> #include <string.h> int main() { int a,b,sum; char str[20]; scanf("%d %d",&a,&b); sum = a + b; sprintf(str,"%d",sum);//字符串地址,第三个参数类型,写入的数据;该函数为清空写入 int len = strlen(str); for (int i=0;i<len;i++) { printf("%c",str[i]); if (str[i] == '-') continue;//第一个字符是负号,则执行下一次循环 if ((i+1)%3 == len%3 && i != len-1) printf(",");//12345678 8位 需要输出为 12,345,678 即 i=1时,2%3==2,8%3==2需要输出‘,’ } return 0; }
参考:
柳婼-https://blog.csdn.net/liuchuo/article/details/54561626
昵称五个字-https://blog.csdn.net/a617976080/article/details/89676670