UVA 10474 - Where is the Marble?

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 next
N lines would contain the numbers written on the N marbles. These marble numbers will not come
in any particular order. Following Q 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 where N = 0 and Q = 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 at y’, if the first marble with number x was found at position y. Positions are numbered
1, 2, … , N.
• ‘x not found’, if the marble with number x 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

题意：现在有n个大理石，并且每个大理石上都写了一个非负整数。要我们找到排序后要查找的元素是否存在.如果存在输出位置。这是一道基础题。应该是为了要我们掌握sort函数和lower_bound检索

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

const int maxn = 10000;

int main()
{
    int n,q,x,a[maxn],kase = 1;
    while(scanf("%d %d",&n,&q),n)
    {
        printf("CASE# %d:\n",kase++);
        for(int i = 0;i < n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        while(q--)
        {
            scanf("%d",&x);
            int p = lower_bound(a,a+n,x) - a;


            if(a[p] == x)
                printf("%d found at %d\n",x,p+1);
            else
                printf("%d not found\n",x);
        }
    }
    return 0;
}