toj 2971 Rotating Numbers
Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 262 Accepted Runs: 86
As CSE students you are already familiar with the binary rotation. Now you should try to do something different like decimal number rotation. It is simple to rotate a decimal number for a single digit to the left to perform left rotation - just place the rightmost digit to the leftmost position as in the following:
12345 => 51234
If we gradually perform the left rotation operation for more and more times we have
51234 => 45123
45123 => 34512 and this way.
But you should be careful to remember that leading zeros in decimal numbers does not carry any significance. Therefore, you need not to add leading zeros.
Input
Each input line consist of an integer M and another integer N. M denotes the number to be left rotated and N indicates how many times to rotate. Both M and N will fit 32-bit integers. Input is terminated by end of file.
Output
The output line contains the number after left rotation is performed on M for N times.
Sample Input
12345 3
1800576 5
10002 5
Sample Output
34512
57618
12
_____________________________________________________________________
Samina Azad (CSE-03)
Source: CUET Easy Contest
#include <string>
#include <cmath>
using namespace std;
int Get(string str)
{
int sum=0;
int i,len=str.length(),k=0;
for(i=len-1;i>=0;i--)
{
sum+=(int)((str[i]-'0')*pow(10.0,k++));
}
return sum;
}
int main()
{
string str;
int n,i,len;
while(cin>>str)
{
cin>>n;
len=str.length();
int num=0;
for(i=0;i<len;i++)
if(str[i]=='0')
num++;
int ss = 0 , kk;
int j=len-1,k;
int t = len - num;
bool mark=false;
for(i=0;i<n;i++)
{
if(num == 0)
{
ss=i;
mark=true;
break;
}
char ch=str.at(j);
str.erase(j);
if(ch!='0')
str=ch+str;
else
{
j--;
num--;
}
}
if (!mark)
printf("%d\n",Get(str));
else
{
kk = n - ss;
ss = kk % t;
for (i = 1 ; i <=ss ; ++ i)
{
char temp=str.at(t-1);
str.erase(t-1);
str=temp+str;
}
printf("%d\n",Get(str));
}
}
return 0;
}
