1001. A+B Format (20) [字符处]
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
题意：计算A+B的和，然后以每三位加个”,” 的格式输出。
分析：把a+b的和转为字符s。除第位是负号的情况，只要当前位的下标i满(i + 1) % 3 == len %
3并且i是最后位，就在逐位输出的时候在该位输出后的后加上个逗号
1002. A+B for Polynomials (25) [模拟]
This time, you are supposed to find A+B where A and B are two polynomials.
Input
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information
of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial,
Ni and aNi (i=1, 2, …, K) are the exponents and coe!icients, respectively. It is given that 1 <= K <= 10，0
<= NK < … < N2 < N1 <=1000.
#include <iostream>
using namespace std;
int main() {
int a, b;
cin >> a >> b;
string s = to_string(a + b);
int len = s.length();
for (int i = 0; i < len; i++) {
cout << s[i];
if (s[i] == '-') continue;
if ((i + 1) % 3 == len % 3 && i != len - 1) cout << ",";
}
return 0;
}
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