1001 A+B Format

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 –​​10⁶≤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

思路一：

      这道题目的意思是计算a+b的和并以千分位的形式输出。注意到a，b的取值范围为–​​10⁶≤a,b≤10​⁶，可以对a，b和sum的大小进行分类讨论,sum的位数最大为7位数，即输出的结果所用的comma最多为两个，所以分三种情况讨论。这种解法有点投机取巧，对a，b的取值有较大的限制。尝试新思路，见思路二。

AC Code：

#include <cstdio>
using namespace std;

void print(int n) {
    if(n < 0) {
        printf("-");
        n = -n;
    }    
    if(n >= 1000000){
        printf("%d,%03d,%03d", n/1000000, (n%1000000)/1000, n%1000);
    }else if(n >= 1000){
        printf("%d,%03d", n/1000, n%1000);
    }else {
        printf("%d", n);
    }
}
int main() {
    int a, b;
    int sum;
    scanf("%d %d", &a, &b);
    sum = a + b;
    print(sum);
    return 0;
} 

思路二：

这道题难点在于如何三位输出，可以利用数组存放每一个数，然后倒序输出，当下标为3的倍数的时候，加入“，”即可。

AC Code：

#include <cstdio>
using namespace std;

void print(int n) {
    if(n < 0) {
        printf("-");
        n = -n;
    }    
    int num[20] = {0};
    int length = 0;
    do{
        num[++length] = n % 10;
        n = n / 10;
    }while(n != 0);
    
    //处理高位 
    int temp = length % 3;
    if(temp) {
        
        for(int i = temp; i > 0; i--) {
            printf("%d", num[length--]);
        }
        if(length > 0)
            printf(",");
    }
    
    int count = 0;
    for(int i = length; i > 0; i--) {
        printf("%d", num[i]);
        count++;
        if(count == 3 && i > 1){
            count = 0;
            printf(",");
        }
    }
}
int main() {
    int a, b;
    int sum;
    scanf("%d %d", &a, &b);
    sum = a + b;
    print(sum);
    return 0;
}