素數 指數 ive test posit muc not ria str

Where is the Marble?

Description

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
分析：這個題目是算法競賽上的一個題目，書中給力詳細的解釋

 1 # include <cstdio>
 2 # include <algorithm>
 3 
 4 using namespace std;
 5 const int maxn = 10000;
 6 int main3()
 7 {
 8     int n, q,a[maxn],kase = 0;
 9     int x;
10     while (scanf("%d %d", &n, &q) == 2 && n)
11     {
12         printf("The NO.%d\n", ++kase);
13         for (int i = 0; i < n; i++)
14         {
15             scanf("%d", &a[i]);
16         }
17             sort(a, a + n);//排序
18             while (q--)//多組輸入
19             {
20                 scanf("%d", &x);//取決於q
21                 int p = lower_bound(a, a + n, x)-a ;//只有指向同一數組的倆個指針變量之間才可以進行計算。否則是沒有意義的。
                                                             兩指針變量相減是兩指針變量相減所得之差，是倆個指針所指數組之間相差的元素個數。實際上是倆個指針值（地址）相減之差再除以該數組元素的長度（字節數），
                                                             註意：因為倆個指針相加沒有任何意義，所以別亂搞。
22                 if (a[p] == x)
23                     printf("%d found at %d\n", x, p + 1);
24                 else
25                     printf("%d not found\n", x);
26 
27                 
28             }
29         
30     }
31     return 0;
32 }


關於第21行lower_bound(p,p+n,x)作用是在p[n]數組中查找大於或等於x的第一個位置，得到的結果是一個指針，前提是數組p[n]進行了排序；
指向同一個數組的兩個指針相減，結果為兩個指針之間的元素數目；
即p[n]存放的第一個數到x這個數之間的元數個數。
這東西有個學名叫叠代器。

排序與檢索【UVa10474】Where is the Marble?