Atcoder 056 D 贪心+二分

D - No Need

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

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.

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Input

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

N Ka1 a2 ... aN

Output

Print the number of the unnecessary cards.

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Sample Input 1

Copy

3 61 4 3

Sample Output 1

Copy

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.

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Sample Input 2

Copy

5 4003 1 4 1 5

Sample Output 2

Copy

5

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

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Sample Input 3

Copy

6 2010 4 3 10 25 2

Sample Output 3

Copy

3

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>using namespace std;int s[5050];bool cmp(int a,int b){return a<b;}int main(){//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);int n,k;scanf("%d%d",&n,&k);int num=n,temp,ans=1;for(int i=1;i<=n;i++){scanf("%d",&temp);if(temp>=k) num--;else s[ans++]=temp;}ans--;sort(s+1,s+num+1,cmp);temp=0;while(num>=2){temp+=s[num];if(s[num-1]+temp<k){num--;continue;}if(s[1]+temp>=k){ans=0;break;}int l=1,r=num-2;while(l+1<r){int mid=(l+r)/2;if(temp+s[mid]>=k) r=mid;else l=mid;}ans=ans-(num-r);num=r;}printf("%d\n",ans);return 0;}