CF Preparing Olympiad (DFS)

最新推荐文章于 2024-11-06 17:30:18 发布

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原文链接：http://www.cnblogs.com/xz816111/p/4572784.html

本文探讨了如何从已制作的问题中为比赛准备一个合适的题目集，包括问题数量、总难度范围、难度差异等限制条件，并提供了求解方法。

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 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

#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <climits>
using    namespace    std;

int    N,L,R,X;
int    ANS,SUM,DIF,NUM,MAX,MIN = 0x7fffffff;
int    S[20];
bool    VIS[20];

void    dfs(int);
int    main(void)
{
    cin >> N >> L >> R >> X;
    for(int i = 0;i < N;i ++)
        cin >> S[i];
    sort(S,S + N);
    dfs(0);
    cout << ANS << endl;

    return    0;
}

void    dfs(int start)
{
    for(int i = start;i < N;i ++)
        if(!VIS[i])
        {
            int    back = DIF;
            int    back_min = MIN;
            int    back_max = MAX;
            MIN = MIN < S[i] ? MIN : S[i];
            MAX = MAX > S[i] ? MAX : S[i];
            VIS[i] = true;
            SUM += S[i];
            DIF = MAX - MIN;
            NUM += 1;
            if(SUM >= L && SUM <= R && NUM >= 2 && DIF >= X)
                ANS ++;
            if(SUM <= R)
                dfs(i);
            VIS[i] = false;
            SUM -= S[i];
            DIF = back;
            NUM -= 1;
            MIN = back_min;
            MAX = back_max;
            DIF = MAX - MIN;
        }
}