HDU 1266 Reverse Number（模拟or数字字符串）

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7836    Accepted Submission(s): 3372

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​​

题解：不断的将数取模取最低位乘10加模。。。。不断循环，直到第一个数为止，尾数遇0则记录个数。。。

AC代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
typedef long long LL;
using namespace std;

int main(){
  int n;
  LL m;
  cin>>n;
  while(n--){
    LL flag=0,t=0;
    int count=0;
    scanf("%lld",&m);
    if(m<0){
      m=-1*m;
      flag=1;
    }
    while(m>0){
      t=t*10+m%10;
      m=m/10;
      if(!t){
        count++; //计算0的个数 
    }    
  }
  if(flag==1){
        cout<<"-";
      }
      cout<<t;
      while(count--)
      {
       printf("0"); 
      }
    printf("\n");
     
  }
  return 0;
}

字符串反向输出：我又没想到。。。。

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
typedef long long LL;
using namespace std;

int main(){
  int t;
  char s[117];
  scanf("%d",&t);
  while(t--)
  {
    scanf("%s",&s);
    int k=0;
      int len = strlen(s);
    if(s[0]=='-')
    printf("-");
    int flag=0;
    for(int i=len-1;i>=0;i--)
    {
      if(s[i]=='0'&&flag==0)
      k++;
      else if(s[i]=='-')
      continue;
      else
      {
        flag=1;
        printf("%c",s[i]);
        
      }
    }
    for(int i=0;i<k;i++)
      printf("0");
      printf("\n");
  }

  return 0;
}