UVA - 10474 Where is the Marble

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

题意:将N个数以升序排序，进行Q次询问，找到输出在第几位，，否则输出not found。

考察sort和lower_bound

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long LL;
int main(){
int N,Q,cas=0;
while(cin>>N>>Q){
if(N==0&&Q==0)
break;
printf("CASE# %d:\n",++cas);
int s[10005];
for(int i=0;i<N;i++)
cin>>s[i];
sort(s,s+N);
while(Q--){
int que,ans;
cin>>que;
ans=lower_bound(s,s+N,que)-s;
if(s[ans]==que)
printf("%d found at %d\n",que,ans+1);
else
printf("%d not found\n",que);
}
}
return 0;
}