排序与检索UVa10474Where is the Marble?

Posted 2020-09-21

                                                                                        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：这个题目是算法竞赛上的一个题目，书中给力详细的解释

 

 

 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--)//多组输入
19             {
20                 scanf("%d", &x);//取决于q
21                 int p = lower_bound(a, a + n, x)-a ;//只有指向同一数组的俩个指针变量之间才可以进行计算。否则是没有意义的。
                                                             两指针变量相减是两指针变量相减所得之差，是俩个指针所指数组之间相差的元素个数。实际上是俩个指针值（地址）相减之差再除以该数组元素的长度（字节数），
                                                             注意：因为俩个指针相加没有任何意义，所以别乱搞。
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 }


关于第21行lower_bound(p,p+n,x)作用是在p[n]数组中查找大于或等于x的第一个位置，得到的结果是一个指针，前提是数组p[n]进行了排序；
指向同一个数组的两个指针相减，结果为两个指针之间的元素数目；
即p[n]存放的第一个数到x这个数之间的元数个数。
这东西有个学名叫迭代器。