关于HDOJ 1266题(Reverse Number) 的理解(C/C++)

    本来应该是一个水题的，结果还是细节上有点小问题，之前WA的2次代码没有处理好中间0，例如输入502，应该输出205的，结果按之前的代码的话就直接给中间0那给截断了，只输出了一个两位数……所以肯定是WA了。之后注意了下中间0，就给AC了。
在这里也给出除了例题的另一组特殊数据吧，方便小伙伴们测试

input：

3
-0012560020
00000
00205



output：

-20065210
0
502

---

题目描述：Reverse Number（数字逆序）

Reverse Number Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

Problem Description
    Welcome to 2006’4 computer college programming contest!

Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one… Ha-Ha!

Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:

1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;
2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;
3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.

Input
    Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.

Output
    For each test case, you should output its reverse number, one case per line.

Sample Input
3
12
-12
1200

Sample Output
21
-21
2100

Author

    lcy

Source

    HDU 2006-4 Programming Contest

原题链接

More info:Question

Accepted代码

#include<iostream>
#include<cmath>
using namespace std;

long long int n,input_;
int main() {
    cin >> n;
    while (n--) {
        cin>>input_;
        long long int x = input_,a,count=0,num=0;
        while (x) {
            a = x % 10;
            x /= 10;
            num++;
        }
        x = input_; a = 10;
        for (int i = 1; i <= num; i++) {
            if (x % a) count++;
            a *= 10;
        }
        long long int res = 0,k=num-1;
        while (count) {
            res += ((long long int)(input_ / pow(10, num - count ) )% 10) * pow(10, k);
            k--; count--;
        }
        cout << res << endl;
    }
    return 0;
}

    ps.直接用int感觉应该也可以过吧，题目里貌似是说了int32位的数，我的AC代码只是因为之前WA时误以为是数据类型的锅，就给改成了long long int，有兴趣的小伙伴可以试试直接跑int。

参考博客

感谢 cnxo   的 HDOJ 1266 Reverse Number（数字反向输出题） 提示了我之前WA考略欠缺的地方（不过貌似他给的数据输出有点不对？已给老哥评论了，等回复）