2013
11-12

# Root of the Problem

Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that AN may be less than, equal to, or greater than B.

The input consists of one or more pairs of values for B and N. Each pair appears on a single line, delimited by a single space. A line specifying the value zero for both B and N marks the end of the input. The value of B will be in the range 1 to 1,000,000 (inclusive), and the value of N will be in the range 1 to 9 (inclusive).

For each pair B and N in the input, output A as defined above on a line by itself.

4 3
5 3
27 3
750 5
1000 5
2000 5
3000 5
1000000 5
0 0

1
2
3
4
4
4
5
16

方法一:
import java.util.*;
import java.io.*;
import java.lang.reflect.Array;

public class Main {
static public void main( String [] string ) throws Exception{
Scanner cin = new Scanner( System.in );
int b, n;
while( true ) {
b = cin.nextInt();
n = cin.nextInt();
if( b == 0 && n == 0 )
break;
int x = (int)Math.pow( b, 1.0/n );
if( Math.abs(Math.pow( x, n )-b)>Math.abs(Math.pow( x+1, n )-b) )
x = x+1;
System.out.println( x );
}
return;
}
}

import java.util.*;

public class Main {

public static void main(String[] args) {
Scanner cin = new Scanner(System.in);
String str[];

while(true)
{
str = cin.nextLine().split(" ");
if(str[0].equals("0") && str[1].equals("0"))
break;

int b = Integer.valueOf(str[0]).intValue();
int n = Integer.valueOf(str[1]).intValue();

int a = findA(b, n);
System.out.println(a);
}
}

private static int findA(int b, int n)
{
int value1, value2 = 0;
int raw1 = 1;
int raw2 = 0;

while(true)
{
raw2 = raw1 + 1;
value1 = (int)(Math.pow(raw1, n));
value2 = (int)(Math.pow(raw2, n));
if(value2 == b)
return raw2;
else if(value2 > b)
break;
else
raw1++;
}

if(Math.abs(value1-b) <= Math.abs(value2-b))
return raw1;
else
return raw2;

}
}

1. /*
* =====================================================================================
*
* Filename: 1366.cc
*
* Description:
*
* Version: 1.0
* Created: 2014年01月06日 14时52分14秒
* Revision: none
* Compiler: gcc
*
* Author: Wenxian Ni (Hello World~), [email protected]
* Organization: AMS/ICT
*
* =====================================================================================
*/

#include
#include

using namespace std;

int main()
{
stack st;
int n,i,j;
int test;
int a[100001];
int b[100001];
while(cin>>n)
{
for(i=1;i>a[i];
for(i=1;i>b[i];
//st.clear();
while(!st.empty())
st.pop();
i = 1;
j = 1;

while(in)
break;
}
while(!st.empty()&&st.top()==b[j])
{
st.pop();
j++;
}
}
if(st.empty())
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}