Codeforces Round #306 (Div. 2) B. Preparing Olympiad dfs

B. Preparing Olympiad

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/550/problem/B

Description

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 Input

3 5 6 1
1 2 3

Sample Output

2

HINT

题意

有n门课，然后让你选择一些课，要求这些课程的和小于等于r，大于等于l，最大值减去最小值至少为x

然后问你有多少种分法

题解：

数据范围小的可怜= =

所以直接dfs吧～

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

ll a[maxn];
int d[20];
int ans=0;
int n,l,r,x;
map<string,int> H;
void dfs(int t,int k[],ll sum,ll mi,ll ma)
{
    if(sum>r)
        return;
    if(ma-mi>=x&&sum<=r&&sum>=l)
    {
        ans++;
    }
    for(int i=t;i<n;i++)
    {
        if(k[i]==0)
        {
            k[i]=1;
            dfs(i,k,sum+a[i],min(mi,a[i]),max(ma,a[i]));
            k[i]=0;
        }
    }
    return;
}
int main()
{
    //test;
    n=read(),l=read(),r=read(),x=read();
    for(int i=0;i<n;i++)
        a[i]=read();
    for(int i=0;i<n;i++)
    {
        d[i]=1;
        dfs(i,d,a[i],a[i],a[i]);
        d[i]=0;
    }
    cout<<ans<<endl;
}