Codeforces Round #306 (Div. 2)B. Preparing Olympiad--状态压缩

You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.

A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.

Find the number of ways to choose a problemset for the contest.

Input

The first line contains four integers n, l, r, x (1 ≤ n ≤ 15, 1 ≤ l ≤ r ≤ 109, 1 ≤ x ≤ 106) — the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 106) — the difficulty of each problem.

Output

Print the number of ways to choose a suitable problemset for the contest.

Sample test(s)

input

3 5 6 11 2 3

output

2

input

4 40 50 1010 20 30 25

output

2

input

5 25 35 1010 10 20 10 20

output

6

Note

In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.

In the second example, two sets of problems are suitable — the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.

In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.

题意：

给你n个数,，还有l,r,x;

让你找一些数的和，满足这些数的和>=l && <=r，并且任意组合中最大值-最小值》=x

求这些组合有多少种；

思路；

这里因为n最多15个，所以我们用状态压缩，例如000111表示取第1,2,3个数；

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<vector>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#define nn 100009
#define inff 0x3fffffff
#define mod 100000000
#define eps 1e-9
using namespace std;
int main()
{
    int i,j;
    int n,l,r,x;
    int a[nn];
    while(scanf("%d %d %d %d",&n,&l,&r,&x)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        int minn=0x3fffffff;
        int maxx=-1;


        int tot=0;
        int sum;
        for(i=0;i<(1<<n);i++)
        {
            sum=0;
            minn=0x3fffffff;
       maxx=-1;
            for(j=0;j<n;j++)
            {
                if(i&(1<<j) )
                {
                    sum+=a[j];
                    minn=min(minn,a[j]);
                    maxx=max(maxx,a[j]);
                }
            }
            if(sum>=l &&sum<=r && maxx-minn>=x)
            {
                tot++;
            }
        }
        printf("%d\n",tot);
    }
}