『NYIST』第九届河南省ACM竞赛队伍选拔赛[正式赛二]--Codeforces -35D. Animals

D. Animals

time limit per test

2 seconds

memory limit per test

64 megabytes

input

input.txt

output

output.txt

Once upon a time DravDe, an outstanding person famous for his professional achievements (as you must remember, he works in a warehouse storing Ogudar-Olok, a magical but non-alcoholic drink) came home after a hard day. That day he had to drink 9875 boxes of
the drink and, having come home, he went to bed at once.

DravDe dreamt about managing a successful farm. He dreamt that every day one animal came to him and asked him to let it settle there. However, DravDe, being unimaginably kind, could send the animal away and it went, rejected. There were exactly n days
in DravDe’s dream and the animal that came on the i-th day, ate exactly ci tons
of food daily starting from day i. But if one day the animal could not get the food it needed, it got really sad. At the very beginning
of the dream there were exactly X tons of food on the farm.

DravDe woke up terrified...

When he retold the dream to you, he couldn’t remember how many animals were on the farm by the end of the n-th day any more, but he
did remember that nobody got sad (as it was a happy farm) and that there was the maximum possible amount of the animals. That’s the number he wants you to find out.

It should be noticed that the animals arrived in the morning and DravDe only started to feed them in the afternoon, so that if an animal willing to join them is rejected, it can’t eat any farm food. But if the animal does join the farm, it eats daily from that
day to the n-th.

Input

The first input line contains integers n and X (1 ≤ n ≤ 100, 1 ≤ X ≤ 104)
— amount of days in DravDe’s dream and the total amount of food (in tons) that was there initially. The second line contains integers ci (1 ≤ ci ≤ 300).
Numbers in the second line are divided by a space.

Output

Output the only number — the maximum possible amount of animals on the farm by the end of the n-th day given that the food was enough
for everybody.

Examples

input

3 4
1 1 1

output

2

input

3 6
1 1 1

output

3

Note

Note to the first example: DravDe leaves the second and the third animal on the farm. The second animal will eat one ton of food on the second day and one ton on the third day. The third animal will eat one ton of food on the third day.

很有意思的一题，因为大家都没有做出来，比赛快结束时才知道要用

freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);    原来是水题，不过思路还是得清晰一点，，简化题意就是如果要容纳这只动物，那么就得吃到最后一天而且不会unhappy，就先把每只动物到最后一天所需的粮食求出来，再排序一下，，，然后贪心策略；

 
   上代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
const int N=500;
int n,m,a[N];
int main()
{
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    int i;
    while(~scanf("%d%d",&n,&m))
    {
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
                          a[i]*=(n-i);//每只动物总共所需的粮食；
        }
        sort(a,a+n);
        printf("\n");
        if(a[0]>m)
            printf("0\n");
        else
        {
            int k=0;
             for(i=0;i<n;i++)
             {
                 if(m-a[i]>=0)
                 {
                     k++;
                     m-=a[i];
                 }
                 else
                    break;
             }
             printf("%d\n",k);
        }
    }
    return 0;
}