PAT (Advanced Level) Practise —— 1001. A+B Format (20)

http://pat.zju.edu.cn/contests/pat-a-practise/1001

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

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

不想写判断语句，用sprintf函数和两个字符串完成：

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>char outStr1[10];char outStr2[10];int main(){int a, b, sum;scanf("%d%d", &a, &b);sum = a + b;if (sum < 0) printf("-");sprintf(outStr1, "%d", abs(sum));int len = strlen(outStr1);int i, j, k;for (i = len-1, j = 0, k = 1; i >= 0; i--, k++){outStr2[j++] = outStr1[i];if (k%3 == 0 && i != 0){outStr2[j++] = ',';}}int len2 = strlen(outStr2);for (i = len2-1; i >= 0; i--){printf("%c", outStr2[i]);}printf("\n");return 0;}

不用字符串，利用printf函数控制输出格式：(http://alvenxuan.iteye.com/blog/1450533)

#include <stdio.h>int main(){int a, b, sum;while (scanf("%d%d", &a, &b) != EOF){sum = a + b;if (sum < 0){sum = -sum;printf("-");}if (sum >= 1000000){printf("%d,%03d,%03d\n", sum/1000000, (sum/1000)%1000, sum%1000);}else if (sum >= 1000){printf("%d,%03d\n", sum/1000, sum%1000);}else{printf("%d\n", sum);}}return 0;}

使用Python：(http://www.cnblogs.com/szhang/archive/2013/01/16/2863511.html)

a,b = raw_input().split()print format(int(a)+int(b),",")

Python在字符处理上真的很强大！(字符串的split函数默认以空格' '作为分割，format函数的第二个参数为","时自动将输出数字以","分割)