PAT 1001. A+B Format (20) c++版

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

题目的意思：给出两个数（-1000000到1000000之间），求它们的和，并按照隔3个输出一个逗号的形式输出。

思路：根据题目意思求出两个数的和，flag来标记正负，将数从个位往前按位转换为char类型的数组，并且每隔三位加入一个逗号。

注意：数字的位数正好是3的整数倍时，第一位上会多出一个逗号。

#include <bits/stdc++.h>using namespace std;int main() {int a, b;cin >> a >> b;int c = a + b;char num[15];int index = 0, time = 0;int flag = 1;if (c < 0) {//如果和是负数，flag记录下，并将它取反flag = 0;c = -c;} else if (c == 0) {//和为0的情况，直接输出并返回cout << "0" << endl;return 0;}while (c > 0) {//index表示要插入的char类型数组的位置num[index++] = '0' + (char)(c % 10);//time用来记录数字的个数time++;c /= 10;//每隔3个数字就加入一个逗号if (time % 3 == 0) num[index++] = ',';}//3个数字加1个逗号，一共4位，index如果是4的整数倍，说明第一位上多了一个逗号if (index % 4 == 0) index--;if (!flag) cout << '-';for (int i = index - 1; i >= 0; i--)cout << num[i];cout << endl;return 0;}