codeforces 306 div.2 B. Preparing Olympiad

B. Preparing Olympiad

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have n problems. You have estimated the difficulty of thei-th one as integerc_i. 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 leastl and at mostr. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at leastx.

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, ..., c_n (1 ≤ c_i ≤ 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.

这题做的比第一题还爽，刚看完子集生成，就碰到这题，一遍AC...

我生成子集的方法是用二进制，每个元素要么在子集中，要么不在，用 j 的二进制形式的每一位代表数组a中对应的位置的元素是否在子集中，在while循环中进行2^n-1次遍历，从而确定所有情况。例如，当i = 5时， j = i = 5，那么j = 0101; 对应的输出 a[0], a[2]，保存在b数组中，记录个数 ，以便往后再判断。

#include<iostream>#include<cstdio>#include<string>#include<algorithm>using namespace std;int main(){int n;long long l, r, x, a[20],b[20];while (cin >> n >> l >> r >> x){int count = 0;for (int i = 0; i < n; i++)cin >> a[i];long long t = 1 << n;long long i, j,k,k1;for ( i = 0; i < t; i++){j = i;k = 0,k1=0;while (j){if (j & 1)b[k1++] = a[k];j >>= 1;++k;}if (k1>1){sort(b, b + k1); int sum = 0;if (abs(b[k1 - 1] - b[0])>=x)for (int m = 0; m < k1; m++){sum += b[m];}if (sum >= l&&sum <= r)count++;}}cout << count << endl;}return 0;}