HDU 1266 Reverse Number (很简单的水题)

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2047    Accepted Submission(s): 926


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
 

Recommend
lxj
 
/*
HDU 1266
水题,但是要用字符串去存储,并不像题目说的是整型的
数据很大
*/
#include
<stdio.h>
#include
<string.h>
int main()
{
char m[100];
long long x;
int T,i;
int t;
scanf(
"%d",&T);
while(T--)
{
scanf(
"%s",&m);
int k=strlen(m);
for(t=k-1;t>=0;t--)
if(m[t]!='0') break;
if(m[0]=='-')
{
printf(
"-");
for(i=t;i>=1;i--)
printf(
"%c",m[i]);
for(i=t+1;i<k;i++)
printf(
"0");
}
else
{

for(i=t;i>=0;i--)
printf(
"%c",m[i]);
for(i=t+1;i<k;i++)
printf(
"0");
}
printf(
"\n");
}
return 0;
}

 

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