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**题目要求:**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解题思路:我们可以将输入的两个值相加后的结果转化为一个字符数组,然后在适当的位置进行输出“,”即可。注意如果结果为负时可先打印负号,然后将结果当作正数讨论较为方便
参考代码:
#include#include #include using namespace std;int main(){ int a,b; cin >>a>>b; int sum = a + b; char result[20] = { 0}; int i=0; //将sum全部转化为正数,这样拆解方便 if(sum<0) { cout <<"-"; sum = -sum; } //当sum为0时特殊讨论 if(sum == 0) { i = 1; result[0] = 0+'0'; } //将sum的各个位拆解并转化成字符保存在数组中 while(sum) { result[i] = sum%10+'0'; sum = sum/10; i++; } for(int j = i-1;j>=0;j--) { cout <
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