PAT 甲级 1001 A+B Format

https://pintia.cn/problem-sets/994805342720868352/problems/994805528788582400

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

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
char s[maxn], out[maxn];
int a[maxn];

void zlr_itoa(int x) {
    if(x == 0) {
        s[0] = '0';
        s[1] = 0;
        return ;
    }
    int top = 0;
    while(x) {
        a[top ++] = x % 10;
        x = x / 10;
    }
    int sz = 0;
    while(top != 0) {
        s[sz++] = (char)(a[-- top] + '0');
        s[sz] = 0;
    }
}

int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    int sum = a + b;
    if(sum == 0) {
        printf("0\n");
        return 0;
    }
    if(sum < 0) {
        printf("-");
        sum = sum * (-1);
    }

    zlr_itoa(sum);
    int len = strlen(s);

    if(len < 3) {
        printf("%s\n", s);
        return 0;
    }

    if(len % 3 == 0) {
        for(int i = 0; i < len - 3; i += 3) {
            for(int j = i; j < i + 3; j ++)
            printf("%c", s[j]);
            printf(",");
        }
        printf("%c%c%c", s[len-3], s[len - 2], s[len - 1]);
    }

    else if(len % 3 == 1) {
        printf("%c,", s[0]);
        for(int i = 1; i < len - 4; i += 3) {
            for(int j = i; j < i + 3; j ++)
                printf("%c", s[j]);
                printf(",");
        }
        printf("%c%c%c", s[len-3], s[len - 2], s[len - 1]);
    }

    else if(len % 3 == 2) {
        printf("%c%c,", s[0], s[1]);
        for(int i = 2; i < len - 5; i += 3) {
            for(int j = i; j < i + 3; j ++)
                printf("%c", s[j]);
                printf(",");
        }
        printf("%c%c%c", s[len-3], s[len - 2], s[len - 1]);
    }

    printf("\n");
    return 0;
}