Codeforces Round #306 (Div. 2) B（dfs)

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 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

1.至少选择两个数；

2.在选择的数中最大的与最小的的差值不小于x；

3.选择的所有数字的总和在l 和 r之间；

简单的dfs，但是很巧妙，搜索前sort一下，用k来记录sum的取舍，k=0表示考虑当前的sum，k=1表示不考虑当前的sum 因为之前已经考虑过它了，这样子就避免了重复。

代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int a[20];int n, l, r, x, cnt;void dfs(int st , int t, int sum , int k){if(!k && sum >= l && sum <= r) cnt++;if(st > t) return;dfs(st+1, t, sum + a[st], 0);    dfs(st+1, t, sum, 1);}int main(){while(~scanf("%d%d%d%d",&n, &l, &r, &x)){for(int i = 0; i < n; i++){scanf("%d", &a[i]);} sort(a, a+n);cnt = 0;for(int i = 0; i < n; i++){for(int j = n-1; j > 0; j--){if(a[j] - a[i] < x || a[j] + a[i] > r) continue;if(j == i+1) {if(a[i] + a[j] >= l) cnt++;}else{dfs(i+1, j-1, a[i] + a[j], 0);}}}printf("%d\n", cnt);}return 0;}