Preparing Olympiad CodeForces #550B 题解[C++]

题目来源

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

题意理解

You have n problems. You have estimated the difficulty of the i-th one as integer c_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 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, …, 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.

Examples

Input

3 5 6 1
1 2 3

Output

Input

4 40 50 10
10 20 30 25

Output

Input

5 25 35 10
10 10 20 10 20

Output

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.

给定 n 道题目, 每道题目都有对应的难度 c. 作为出题人, 你被要求给出一个题目组合, 其中
1. 总难度必须在 [l, r] 的范围内;
2. 至少含有两道题;
3. 最简单题和最难题的难度差距必须>=x;

输入n, l, r, x以及每道题对应的难度, 输出有多少种可能的组合.

解题思路

这道题一看看过去就是一个暴力搜索, 可以考虑位掩码的思想. 因为最多只有15道题目, 所以可以很容易确定题目组合情况的上限即是: 1<<15.

之后我们依次遍历搜索每一种情况, 判断该种情况是否符合要求即可. 而其中遍历搜索的部分便是求解此题的核心. 这里就要用到位掩码的思想进行遍历.

构造一个15 bit的数, 我们先把这个数赋值为0, 每次迭代进行增1代表对这种情况进行搜索. 然后因为我们总共只有15道题, 每一bit就代表是否要把这个bit对应的题加入到我们的试卷中. 这里可以利用位运算进行快速的判断, 确定该bit为0还是为1.

根据上面的思想, 我们很容易地就能写出代码如下.

代码实现

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int mask[15];
int ans = 0;
int n, l, r, x, temp;

int main()
{

    long ans = 0;
    scanf("%d%d%d%d", &n, &l, &r, &x);
    vector<int> problems;
    for (int i = 0; i < n; i++) {
        scanf("%d", &temp);
        problems.push_back(temp);
    }

    int total_case = 1 << n;
    for (int i = 0; i < total_case; i++) {
        int mi = INT_MAX;
        int ma = INT_MIN;
        int sum = 0;
        for (int j = 0; j < n; j++) {
            if (i&(1 << j)) {
                mi = min(mi, problems[j]);
                ma = max(ma, problems[j]);
                sum += problems[j];
            }
        }
        if (sum >= l && sum <= r && ma - mi >= x) ans++;
    }

    printf("%d\n", ans);

    //system("PAUSE");
    return 0;

}