一、Question

1. 题目描述

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.

2. 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.

3. 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:
（1） ‘x found at y’, if the first marble with number x was found at position y. Positions are numbered
1, 2, . . . , N.
（2） ‘x not found’, if the marble with number x is not present.
Look at the output for sample input for details.

4. Sample Output

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

5. Sample Output

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

二、题解

1. C++

#include
using namespace std;int main()
{int m,n;int cnt = 0;while (scanf("%d %d", &m ,&n) == 2){if(m==0&&n==0)break;++cnt;int marble[m];for(int i = 0; i < m; ++i){scanf("%d", &marble[i]);}int query[n], pos[n];memset(query, 0, n);memset(pos, 0, n);sort(marble, marble + m);for(int j = 0; j < n; ++j){scanf("%d", &query[j]);int k ;for(k = 0; k < m; ++k){if(query[j] == marble[k]){int tmp = k + 1;pos[j] = tmp;break;}}if(k == m){pos[j] = 9999;}}printf("CASE# %d:\n",cnt);for(int p = 0; p < n; ++p){if(pos[p]==9999){printf("%d not found\n", query[p]);}else{printf("%d found at %d\n", query[p], pos[p]);}}}
}

其实，我们利用C++的STL库可以更加便捷地解决问题，lower_bound可以节省一个for循环和两个数组的开销，代码如下：

#include
using namespace std;
const int maxn = 100000;int main()
{int m, n, x, a[maxn], cnt = 0;while (scanf("%d%d", &m, &n)==2 && m){printf("CASE# %d:\n", ++cnt);for(int i = 0; i < m; ++i)scanf("%d", &a[i]);sort(a, a+m);while (n--){scanf("%d", &x);int p = lower_bound(a, a+n, x) - a; //在已排序数组a中寻找xif(a[p] == x)printf("%d found at %d\n", x, p+1);elseprintf("%d not found \n", x);}}return 0;
}

2. Python

if __name__ == "__main__":cnt = 0while True:m, n = map(int, input().split())if m == 0 and n == 0:breakcnt = cnt + 1marble = [0] * mfor i in range(m):marble[i] = int(input())query = [0] * npos = [0] * nmarble.sort()for j in range(n):query[j] = int(input())k = 0ttt = 0for k in range(m):if query[j] == marble[k]:tmp = k + 1pos[j] = tmpbreakelse:ttt = k + 1if ttt == m:pos[j] = 9999print('CASE# %d:\n' % cnt)for p in range(n):if pos[p] == 9999:print('%d not found\n' % query[p])else:print("%d found at %d\n" % (query[p], pos[p]))

注：代码中需要判定 这个循环是否跑到了最后都没有执行break；不可以直接在后面判定 ，因为python和C++的for循环是不同的，for k in range(m)最后k执行到m-1后便自动退出了，然而C++的for循环最后还要执行一个++操作，知道k不小于m为止，循环结束后k等于m，这点需要提起注意。