排序检索C - Where is the Marble?
2015-08-18 16:15
253 查看
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 nd the rst 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
xed 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 rst 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.
SampleInput
4 1
2
3
5
1
5
5 2
1
3
3
3
1
2
3
0 0
SampleOutput
CASE# 1:
5 found at 4
CASE# 2:
2 not found
3 found at 3
要找的数字在最下面
输入的几个数字在查找之前要先排序
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 nd the rst 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
xed 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 rst 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.
SampleInput
4 1
2
3
5
1
5
5 2
1
3
3
3
1
2
3
0 0
SampleOutput
CASE# 1:
5 found at 4
CASE# 2:
2 not found
3 found at 3
要找的数字在最下面
输入的几个数字在查找之前要先排序
#include <stdio.h> #include <stdlib.h> #include <string.h> int cmp(const void *a, const void *b){ return (*(int *)a- *(int *)b); } int main(){ int n, q, times = 1, qnum, nnum[10010]; while (scanf("%d%d", &n, &q) != EOF) { if (n == 0 && q == 0) break; memset(nnum, 0, sizeof(nnum)); printf("CASE# %d:\n", times++); for (int i = 0; i < n; i++) scanf("%d", &nnum[i]); qsort(nnum, n, sizeof(nnum[0]), cmp); for (int i = 0; i < q; i++) { scanf("%d", &qnum); int mark = 0, flag = 0; for (int j = 0; j < n; j++) if (qnum == nnum[j]) { printf("%d found at %d\n", qnum, j + 1); flag = 1; break; } if (flag == 0) printf("%d not found\n", qnum); } } return 0; }
相关文章推荐
- 阅读方法
- Linux 内核调试
- FastDFS在.Net平台上的使用
- Linq To EF
- struts2 json返回试验
- 所有中心对称五字母域名生成,扫了一下,com的基本上都被注册了。。。
- 第六周 G题
- ziji
- jquery-chosen 选择框插件
- java.util.date类型保存到mysql数据库报错的问题
- Ubuntu上编译安装Kamailio
- oracle-创建表空间报错 提示ora-01119 ora-27040:无法创建文件
- POJ 2259 Team Queue 数据结构 队列
- activemq实现简单的消息传递(java 实现)
- Gas Station [leetcode] 两个解决方案
- hdu1570 水~水~水~
- i18n
- Java面试题大全
- 浅谈脱壳中的Dump技术
- NVMe:PCIe SSD标准不断完善,直指Fusion-io