#include <iostream>
#include <algorithm>
using namespace std;
main()
{
int n, q, x, a[10000], k = 0;
while(cin >> n >> q)
{
cout << "CASE#" << k << ":" << endl;
k++;
for(int i = 0; i < n; i++)
{
cin >> a[i];
}
sort(a, a+n);
while(q--)
{
cin >> x;
int p = lower_bound(a, a+n, x) - a; //在已知数组a中寻找x
if(a[p] == x)
cout << x << " found at " << p << endl;
else
cout << x << " not found" << endl;
}
}
}
题目
现有N个大理石,每个大理石上写了一个非负整数、首先把各数从小到大排序然后回答Q个问题。每个问题问是否有一个大理石写着某个整数x,如果是,还要
回答哪个大理石上写着x。排序后的大理石从左到右编号为1~N。(在样例中,为了
节约篇幅,所有大理石的数合并到一行,所有问题也合并到一行。)
样例输入:
4 1
2 3 5 1
5
5 2
1 3 3 3 1
2 3
样例输出:
CASE# 1:
5 found at 4
CASE# 2:
2 not found
3 found at 3