1001 A+B Format

Posted 2021-03-01 trebienba

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 −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

审题

题目给出两个整数a,b（ −10^6≤a,b≤10^6），计算出它们的和sum,然后每三位用逗号隔开输出。

思路

因为a,b都没有超过int类型的范围，所以可知直接定义int类型计算。根据a,b的取值范围，可以得知sum不会超过10^7。因此可以把结果分为三类:

1.sum的位数大于6.

printf("%d,%03d,%03d", sum / 1000000, sum % 1000000 / 1000, sum % 1000);

2.sum的位数大于3小于6.

printf("%d,%03d", sum / 1000, sum % 1000);

3.sum的位数小于3。

printf("%d", sum);

参考代码

 1 #include<stdio.h>
 2 ?
 3 int main(){
 4   int a, b, sum;
 5   scanf("%d %d", &a, &b);
 6   sum = a + b;
 7   if(sum < 0){
 8     printf("-");
 9     sum = -sum;
10   }
11   
12   if(sum >= 1000000){      //当位数大于6时
13     printf("%d,%03d,%03d", sum / 1000000, sum % 1000000 / 1000, sum % 1000);
14   }
15   else if(sum >= 1000){    //当位数大于3小于6时
16     printf("%d,%03d", sum / 1000, sum % 1000);
17   }
18   else{                    //当位数小于等于3时
19     printf("%d", sum);
20   }
21   return 0;
22 }