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

 

题目大意是找到指定数字（经过排序后的）位置，用sort()和lower_bound()就可以做了，记住的是不能用upper_bound().

贴代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 10005;
int a[maxn],b[maxn];

bool cmp(int a,int b){
    return a < b;
}
int main()
{
    int n,q;
    int kase = 1;
    while(cin>>n>>q){
        if(n == 0 || q == 0) break;
        for(int i = 0;i < n;i++){
            cin>>a[i];
        }
        for(int i = 0;i < q;i++){
            cin>>b[i];
        }
        cout<<"CASE# "<<kase++<<":"<<endl;
        sort(a,a+n,cmp);
        for(int i = 0;i < q;i++){
            int t = lower_bound(a,a+n,b[i]) - a;
            if(a[t] == b[i]){
                cout<<b[i]<<" found at "<<t+1<<endl;
            }
            else{
                cout<<b[i]<<" not found"<<endl;
            }
        }
    }
    return 0;
}