HDU 1266 Reverse Number （water ver.）

HDU 1266 Reverse Number （water ver.）2014-10-04 csdn synapse7

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

312-121200

Sample Output

21-212100

水。

完整代码：

/*0ms,208KB*/#include<cstdio>#include<cstring>char s[15];void solve（char* s）{int i, j, len = strlen（s）;for （i = len - 1; i > 0; --i）if （s[i] ！= "0"） break;for （j = i; j >= 0; --j） putchar（s[j]）;puts（s + i + 1）;}int main（）{int t;scanf（"%d
", &t）;while （t--）{gets（s）;if （s[0] == "-"）{putchar（"-"）;solve（s + 1）;}else solve（s）;}return 0;}

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