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1411051730-hd-Reverse Number

2014-11-05 18:10:00.0 hdoj  
导读:Reverse NumberTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5222    Accepted Submission(s): 2435Problem DescriptionWelcome to 2006'4 compute...。。。


Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5222    Accepted Submission(s): 2435


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
 题目大意
        给一个数字按照规则逆序输出。
解题思路
       一开始输入方式是int型的时候超时了,哎,,,,,又没有超过int32位,用得着char字符型吗?纠结,好吧,想要ac不超时输入方式用char型。
代码
#include#includechar num[110];int main(){ int t; int i,j,k,l,len; scanf("%d",&t); getchar(); while(t--) {  scanf("%s",num);  len=strlen(num);  k=0;  if(num[0]=='-')  {   k=1;//这是第一点要求    printf("-");  }  j=0;  for(i=len-1;i>=k;i--)  {   if(num[i]=='0')       j++;   else       break;  }//这是第三点要求   for(l=i;l>=k;l--)//第二点要求       printf("%c",num[l]);  for(l=0;l

(编辑: wangluoershixiong)

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