Where is the Marble?

Posted 2020-08-29

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 di?erent 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

 

题意： 输入数 N和Q   N代表你要输入的数有多少个， Q是你要查询的数有多少个，输入完N 输入Q（好几个），找你输入是否存在，存在就输出他所在的位置。

 方法 ：先排序, 用 lower_boundf返回可插入的位置的下标，然后判断这个下标的数， 是否等于要查询的数，等于就输出，不等于就 no found。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int N , Q;
    int i, L = 0;
    int a[10000], b[10000];

    while(scanf("%d%d",&N,&Q)!=EOF)
    {
        if(N == 0 && Q == 0)
            break;
        memset( a, 0, sizeof(a));
        memset( b, 0, sizeof(b));
        int t = 0,flag = 0 , d, q;
        q = Q;

        for( i = 0; i < N; i++)
            scanf("%d",&a[i]);
        sort(a,a+N);
        while(Q--)
            { scanf("%d",&b[t++]); }
        printf("CASE# %d:\n",++L);
        for( i =0 ;i < q; i++)
          {
              d = lower_bound(a,a+N,b[i])-a;
              if(a[d]==b[i])
              printf("%d found at %d\n",b[i],d+1);
              else
              printf("%d not found\n",b[i]);
          }



    }
}