Problem Description

AtCoDeer the deer has N cards with positive integers written on them. The number on the i-th card (1≤i≤N) is ai. Because he loves big numbers, he calls a subset of the cards good when the sum of the numbers written on the cards in the subset, is K or greater.

Then, for each card i, he judges whether it is unnecessary or not, as follows:

If, for any good subset of the cards containing card i, the set that can be obtained by eliminating card i from the subset is also good, card i is unnecessary.
Otherwise, card i is NOT unnecessary.
Find the number of the unnecessary cards. Here, he judges each card independently, and he does not throw away cards that turn out to be unnecessary.

Constraints

All input values are integers.

1≤N≤5000

1≤K≤5000

1≤ai≤109(1≤i≤N)

Partial Score

300 points will be awarded for passing the test set satisfying N,K≤400.

Input

The input is given from Standard Input in the following format:

N K
a1 a2 ... aN

Output

Print the number of the unnecessary cards.

Example

Sample Input 1

3 6
1 4 3

Sample Output 1

1
There are two good sets: {2,3} and {1,2,3}.

Card 1 is only contained in {1,2,3}, and this set without card 1, {2,3}, is also good. Thus, card 1 is unnecessary.

For card 2, a good set {2,3} without card 2, {3}, is not good. Thus, card 2 is NOT unnecessary.

Neither is card 3 for a similar reason, hence the answer is 1.

Sample Input 2

5 400
3 1 4 1 5

Sample Output 2

5
In this case, there is no good set. Therefore, all the cards are unnecessary.

Sample Input 3

6 20
10 4 3 10 25 2

Sample Output 3

3

题意：给出 n 个数，对于这个数组中的任意一个子集，如果他的元素和大于等于 k，那么这个子集就是一个好的子集，如果将一个数所在的所有的好的子集都删去这个数，但这些集合仍是好的子集，那么这个数就称为无用数，问这 n 个数中无用数的个数

思路：

首先，如果一个数不是无用数，由于若在子集中删除大于等于他的数后无法保证集合仍是好的子集，因此所有大于等于这个无用数的数都不是无用数

根据以上推论，首先将数组进行排序，然后二分即可，对于 mid>=k 的数，这些数肯定不是无用数，而对于 mid<k 的数，则需要将这些数去掉

在将 mid<k 的数去掉的过程中，需要去在数组中判断构成的集合的和是否存在属于 [k-a[mid],k-1] 的数，在这个过程中，暴力求解时间复杂度会达到 O(n^3)，由于 N 最大到 5000，那么从整体的复杂度来说会超时，因此在求和过程需要进行优化

考虑使用二进制优化，利用 bitset 容器，每次用 sum|=sum<<a[i] 来求和，最后枚举一遍，看 [sum[k-a[mis]],sum[k-1]] 中是否存在 1 即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 5000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;int n,k,a[N];
bitset<N> sum;
bool judge(int x) {if(a[x]>=k)return 1;sum.reset();sum[0]=1;for(int i=1;i<=n;i++)if(i!=x)sum|=sum<<a[i];for(int i=k-1;i>=k-a[x];i--)if(sum[i])return true;return false;
}
int main() {scanf("%d%d",&n,&k);for(int i=1;i<=n;i++)scanf("%d",&a[i]);sort(a+1,a+n+1);int left=0,right=n+1;while(left<right) {int mid=(left+right)>>1;if(judge(mid))right=mid;elseleft=mid+1;}printf("%d\n",left-1);return 0;
}

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No Need（AtCoder-2346）

Problem Description

Constraints

Partial Score

Input

Output

Example

Source Program

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