[PAT 甲级] 1001 A+B Format （20 分)

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1001 A+B Format （20 分)

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 −10
​6
​​ ≤a,b≤10
​6
​​ . 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
Sample Output:
-999,991

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这一题关键的就是加逗号，有两种方法一种是从前往后加逗号，一种是从后往前加逗号，个人比较喜欢从前往后，因为不用特判负数

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从前往后，满足(i + 1) % 3 == len % 3;特判最后一位。

#include<iostream>
#include<cmath>
using namespace std;
int main () {
    int a, b;
    cin >> a >> b;
    string c = to_string(a + b);
    int len = int(c.length());
    string ans = "";
    for (int i = 0; i <= len - 1; i++) {
        ans += c[i];
        if (c[i] == '-') {
            continue;
        }
        if ((i + 1) % 3 == len % 3 && i != len - 1) {
            ans = ans + ',';
        }
    }
    cout << ans;
    return 0;
}

从后往前，满足 num % 3 == 0,特判0。

#include<iostream>
#include<cmath>
using namespace std;
int main () {
    int a, b, c, sign = 0;
    cin >> a >> b;
    c = a + b;
    string ans;
    if (c < 0)  sign = 1;
    if (c == 0) ans = "0";
    c = abs(c);
    int num = 1;
    while(c != 0) {
        ans = char(c % 10 + '0') + ans;
        c /= 10;
        if (num++ % 3 == 0 && c != 0) {
            ans = ',' + ans;
        }
    }
    if (sign == 1) {
        ans = '-' + ans;
    }
    cout << ans;
    return 0;
}