UVa 10474 Where is the Marble?
2016-05-25 14:45
295 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.
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
解析:排序和查找,简单题。运用STL很容易解决。
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.
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
解析:排序和查找,简单题。运用STL很容易解决。
#include <cstdio> #include <algorithm> using namespace std; const int MAXN = 10000; int a[MAXN+10]; int main() { int n, q, x, cn = 0; while(scanf("%d%d", &n, &q), n){ printf("CASE# %d:\n", ++cn); for(int i = 0; i<n; ++i) scanf("%d", &a[i]); sort(a, a+n); while(q--){ scanf("%d", &x); int idx = lower_bound(a, a+n, x)-a; if(a[idx] == x) printf("%d found at %d\n", x, idx+1); else printf("%d not found\n", x); } } return 0; }
相关文章推荐
- jQuery插件
- apusic7配置2
- 测试下载
- shell之入门篇
- 328. Odd Even Linked List #Medium
- 剑指Offer——之字形打印二叉树
- 边长、边数可配置的旋转多面体
- Java泛型中遇到的协变问题
- select count(*)
- 实现选择排序算法
- MySQL无法存储Emoji表情问题
- DMA
- iOS开发----懒加载
- Eclipse调试技巧总结
- IO-同步异步,阻塞非阻塞,select, poll , epoll
- 怎么利用GitHub
- 如何判断listview的item已经滑出或者滑入屏幕
- linux系统中安装svn
- QT与JavaScript互调 javaScriptWindowObjectCleared()信号
- 使用Eclipse RCP创建视图并实现视图间消息传递