CodeForces 387B George and Round

B. George and Round

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

George decided to prepare a Codesecrof round, so he has prepared m problems for the round. Let's number the problems with integers 1 throughm.
George estimates the i-th problem's complexity by integer bi.

To make the round good, he needs to put at least n problems there.
Besides, he needs to have at least one problem with complexity exactly a1,
at least one with complexity exactly a2,
..., and at least one with complexity exactly an.
Of course, the round can also have problems with other complexities.

George has a poor imagination. It's easier for him to make some already prepared problem simpler than to come up with a new one and prepare it. George is magnificent at simplifying problems. He can simplify any already prepared problem with complexity c to
any positive integer complexity d(c ≥ d),
by changing limits on the input data.

However, nothing is so simple. George understood that even if he simplifies some problems, he can run out of problems for a good round. That's why he decided to find out the minimum number of
problems he needs to come up with in addition to the m he's prepared in order to make a good round. Note that George can come up with
a new problem of any complexity.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 3000)
— the minimal number of problems in a good round and the number of problems George's prepared. The second line contains space-separated integers a1, a2, ..., an (1 ≤ a1 < a2 < ... < an ≤ 106)
— the requirements for the complexity of the problems in a good round. The third line contains space-separated integers b1, b2, ..., bm (1 ≤ b1 ≤ b2... ≤ bm ≤ 106)
— the complexities of the problems prepared by George.

Output

Print a single integer — the answer to the problem.

Sample test(s)

input

3 5
1 2 3
1 2 2 3 3

output

0

input

3 5
1 2 3
1 1 1 1 1

output

2

input

3 1
2 3 4
1

output

3

Note

In the first sample the set of the prepared problems meets the requirements for a good round.

In the second sample, it is enough to come up with and prepare two problems with complexities 2 and 3 to
get a good round.

In the third sample it is very easy to get a good round if come up with and prepare extra problems with complexities: 2, 3, 4.

思路：很简单的贪心题目，不过一开始理解错题意了，题目意思是找至少满足n个条件的数，给出的b数组的数据只要满足bi>=aj，该数就可以单做满足条件的数，遍历一下就行了。

AC代码如下：

#include <iostream>
#include <cstring>
#include <set>
using namespace std;

const int maxn=3000+5;
int a[maxn];
int b[maxn];

int main(){

    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++)
        cin>>a[i];
    for(int i=0;i<m;i++){
        cin>>b[i];
    }
    int ans=0;
    for(int i=0,j=0;i<n && j<m;){
        if(a[i]<=b[j]){
          ans++;i++,j++;
        }
        else j++;
    }
    cout<<n-ans<<endl;

    return 0;
}