amazon 2013校园招聘题目

　　下面是amazon 2013校园招聘题目，望大家贴现自己的答案哈，我准备收集笔试面试题。

Amazon Hiring Campus 2013 - Final 6
consider a kind of configuration file in amazon sofeware system. this kind of configuration file's format looks like this:
B=10;
A={
  A=100;
  B=BDE;
  C=C;
  D={
     A=windows;
     B=mac;
     C={
       A=redhat;
       B=ubuntu;
      };
 };
A+={
 A=200;
 E=1000;
};
to repsent the key of the configuration, we use period(.) delimated method. for example,A.B represents the element B in Map A, and  the value of A.B is BDE; similarly,  the value of A.D.C.A is redhat. the the represent string is called 'path expression'. 
the configuration also support append and override operation. for the above example, we use += operation to append or override the value in Map A. now the value of A.A is 200, and the value of A.E is 1000;
Now, given a configuration strings and the key path of configuration, please return the value of configuration based the configuration strings. 

 Rules
1)  the key name and his value only contains alphabet(A-Z,a-z) and number(0-9), no other charcters; 
2) if cannot find the value or the expression point to a map, please output "N/A"
3) if find the value, please output the value. no spaces when output the value. 
 
Input and Output
there are three part sin the input. the first line contains two integers indicates  the number of congiruation lines(M) and the number of expressions(N).
M<=100, N<=100. the following M lines are the confugration and the last N lines are expression. every configuration line contains one or more configurations. every line length less than 1000. 
Input :
2 2
A={A=1;B=2;C=3;E={A=100;};};
A+={D=4;E={B=10;C=D;};};
A.E.B
B.D.E
Output 
A.E.B=10
B.D.E=N/A





#include <stdio.h>
#include <string.h>
#include <malloc.h>
void calculateAndPrint(char* confs[], int confLength, char* exs[] , int exLength){
    //Your Code is here
}
int main(){
    int confLength=0; 
    int exLength=0;
    scanf("%d %d\n",&confLength, &exLength);
    char* confs[10];
    char* exs[10];
    int i=0;
    for(i=0;i<confLength;i++){
        char* data = (char*)malloc(1000*sizeof(char));
        gets(data);
        confs[i] = data;
    }
    for(i=0;i<exLength;i++){
        char* data = (char*)malloc(1000*sizeof(char));
        gets(data);
        exs[i] = data;
    }
    calculateAndPrint(confs,confLength,exs,exLength);
    for(i=0;i<confLength;i++){
        free(confs[i]);
    }
    for(i=0;i<exLength;i++){
        free(exs[i]);
    }
    return 0;
} 




•    Question 1 / 2
Let's assume that there is a simple market for beans. Every day there is a published bean price in the market. Traders can buy or sell at the published price. There is a trader who time travelled to future and brought back the price information for a number of days in the future. If you have this information and you are allowed to buy and sell many times. How do you make the maximum profit? The price information will be given as an array of numbers. Each number is for a day’s trading price. The numbers are all integers to simplify the problem. You will need to return the index of the buy-in point and sell-out point for maximum profit.
Rules:
1) The input line length less than 1000, and the trading price length less than 100;
2) The trading price is positive integer;
3) The trading prices are delimited  by ' '(single space);
4) Please make sure every buying and selling period shortest. especially, please ouput '-' if all the trading prices are the same or no trading point;
Sample Input and Output:
Input 1
1 3 5 4 2 8 10
Output 1
1 3 5 7
To make the maximum profit, you should buy at $1 and sell at $5, and then buy at $5 and sell it at $10. so the output is "1 3 5 7".
Input 2 
1 1 1 3 5 4 2 2 2 8 10
Ouput 2
3 5 9 11 
 

include <iostream>
#include <cstring>
 
using namespace std;
 
 
void calculateAndPrint(int array[], int length)
{
    //Your Code is here
    const int halflen = length / 2;
    int buy_index[halflen];
    int sell_index[halflen];
    memset(buy_index, 0, halflen);
    memset(sell_index, 0, halflen);
    
    int sum = 0;  
 
    int b = 0;  
    
    int i = 0;
    int j = 0;
    int k = 0;
    for(; i < length; i = j +1)  
    {  
        for(j = i; i <length; j++)
        {
            b = array[j] - array[i];
            if(b < 0) 
            {
                b = array[i];  
            }
            else
            {
                b += array[i]; 
            }
            if(sum < b) 
            {
                sum = b; 
                buy_index[k] = i;
                sell_index[k] = j;
                k++;
            }
        }
    } 
 
    if(k >0)
    {
        for(i = 0; i < k; i++)
        {
            cout<<buy_index[k]<<" "<<sell_index[k]<<" ";
        }
        cout<<endl;
    }
    else
    {  
        cout<<"-";
    }
}
 
int splitAndConvert(char* strings,int array[])
{
    char*tokenPtr = strtok(strings," ");
    int i=0;
    while(tokenPtr!=NULL)
    {
        array[i] = atoi(tokenPtr);
        i++;
        tokenPtr=strtok(NULL," ");
    }
    return i;
}
 
int main()
{
    char line[1000] = {0} ;
    while(gets(line))
    {
        int array[100] = {0};
        int length = splitAndConvert(line,array);
        if(length==0)
        {
            break;
        }
        calculateAndPrint(array, length);
        cout<<endl;
    }
    return 0; 
}