uvaoj-12474：大理石在哪裏

Raju and Meena love to play with Marbles. They have got a lot of marbles with numbers written on them. At the beginning, Raju would place the marbles one after another in ascending order of the numbers written on them. Then Meena would ask Raju to find the first marble with a certain number. She would count 1...2...3. Raju gets one point for correct answer, and Meena gets the point if Raju fails. After some fixed number of trials the game ends and the player with maximum points wins. Today it's your chance to play as Raju. Being the smart kid, you'd be taking the favor of a computer. But don't underestimate Meena, she had written a program to keep track how much time you're taking to give all the answers. So now you have to write a program, which will help you in your role as Raju.

 

Input 

There can be multiple test cases. Total no of test cases is less than 65. Each test case consists begins with 2 integers:N the number of marbles and Q the number of queries Mina would make. The nextN lines would contain the numbers written on theN marbles. These marble numbers will not come in any particular order. FollowingQ lines will have Q queries. Be assured, none of the input numbers are greater than 10000 and none of them are negative.

Input is terminated by a test case whereN = 0 andQ = 0.

 

Output 

For each test case output the serial number of the case.

For each of the queries, print one line of output. The format of this line will depend upon whether or not the query number is written upon any of the marbles. The two different formats are described below:

 

• `x found aty', if the first marble with numberx was found at position y. Positions are numbered1, 2,..., N.
• `x not found', if the marble with numberx is not present.

Look at the output for sample input for details.

 

Sample Input 

4 1
2
3
5
1
5
5 2
1
3
3
3
1
2
3
0 0

 

Sample Output 

CASE# 1:
5 found at 4
CASE# 2:
2 not found
3 found at 3

題解：題不難，主要是c++STL的運用，今天沒發博客，總感覺怪怪的，正好新學到了STL，來mark下；

code：

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    int num_q,num_n;
    int a[10005];
    int times=0;
    while(cin>>num_n>>num_q&&num_n)
    {
        cout<<"CASE# "<<++times<<':'<<endl;
        for(int i=0; i<num_n; i++)
        cin>>a[i];
        sort(a,a+num_n);
        int ans;
        int x;
        while(num_q--)
        {
            cin>>x;
            ans=lower_bound(a,a+num_n,x)-a;
            if(a[ans]==x) cout<<x<<" found at "<<ans+1<<endl;//注意要加一哦，因爲數組空間是從零開始的，而大理石卻是從一開始數的；
            else cout<<x<<" not found"<<endl;
        }
    }
    return 0;
}

筆記：sort函數，可以自己編寫cmp函數來控制，也可以利用cmp對結構體排序：

code：

struct Time
{
   int a;
   int b;   
  }
  bool compare(Time A,Time B)
  {
      return A.b<B.b;        
}         

情人節快樂！~