PAT甲级 1001. A+B Format (20)

1001. A+B Format (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:把数字格式化，每三个数字中间要加上逗号。
思路：水题,数字小于等于三个就不需要加逗号了。
AC代码：

#include<iostream>
#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<bitset>
#include<string>
using namespace std;
const int N_MAX = 1000000;
string transform(int a) {
  string s ="";
  if (a == 0) { s += '0'; return s; }
  int k = 0;
  while (a) {
    s+= '0'+(a % 10);
    a /= 10;
  }
    return s;
}

string judge(string s,bool flag) {

  if (s.size() <= 3) { 
    if (!flag) {
      reverse(s.begin(), s.end());
      return s;
    }
  else {
    s += '-';
    reverse(s.begin(), s.end());
    return s;
     }
  }
  string cur;
  int size = s.size(),k=0;
  
  while (size > 3) {
    cur += s.substr(k,3);
    cur += ',';
    k += 3;
    size -= 3;
  }
  if (size <= 3 && size > 0)cur += s.substr(k, s.size());
  if (flag)cur += '-';
  reverse(cur.begin(), cur.end());
  return cur;
}


int main() {
  int a, b;
  bool flag;
  while (cin>>a>>b) {
    flag = 0;//判断正负
    int c = a + b;
    if (c < 0)flag = 1;
    c = abs(c);
    string s= transform(c);
    s = judge(s,flag);
    cout << s << endl;
  }
  return 0;
}