题目详情 - 1001 A+B Format (pintia.cn)
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
- 1
Sample Output:
-999,991
- 1
a+b 的结果转换成字符串,方便处理,这里用到的库函数是 to_string(sum) ,它可以将数字 sum 转换成字符串。x 中,从后往前进行遍历,每次遍历都将 x[i] 加入到答案字符串数组 res 中。注意是 res = x[i] + res ,因为是从后往前遍历,所以每次加入的数字要放在 res 左边。j 取余来判断当前是否已经遍历了 3 个数字,除此之外还要注意遍历到第一个数字的情况,如果已经遍历到第一个数字,即使已经遍历了 3 个数字也不能加逗号。#include
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
//进行字符串转化
string x = to_string(a + b);
//从后往前遍历
string res = "";
for (int i = x.size() - 1, j = 0; i >= 0; i--)
{
j++;
res = x[i] + res;
//每遍历三个数字加一个逗号,但是要排除作为第一个数字的情况
if (j % 3 == 0 && i && x[i - 1] != '-') res = "," + res;
}
cout << res << endl;
return 0;
}