codeforces 550B Preparing Olympiad(DFS+回溯)

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 1
1 2 3

output

2

input

4 40 50 10
10 20 30 25

output

2

input

5 25 35 10
10 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.

题目大意：要求找一个题目组，题目不能少于2题，每个题目都有相应的难度，满足2个条件：1、要求找到的题目的总难度要在l,r之间。2、最简单的和最难的题目的差至少要大于等于X 。

#include<stdio.h>

#include<string.h>

#include<iostream>

#include<algorithm>

using namespace std;

int c[20],t,vis[20],n,l,r,x,m,p[20];

void dfs(int y,int len,int cnt,int sum,int k) // y代表当前是哪个题目了，len表示要选的题目数量，cnt代表当前已选的题目数量，sum代表题目的总难度，k代表已经存的题目（p数组存放题目的难度，用于判断最简单和最难的题目的难度差）

{

if(sum>r)return;

if(sum<l&&cnt==len)return;

if(sum>=l&&sum<=r&&cnt==len&&p[k-1]-p[0]>=x){ m++;return;}

for(int i=y;i<=n;i++)

{

if(!vis[i]){

vis[i]=1;

p[k]=c[i];

dfs(i,len,cnt+1,sum+c[i],k+1);

}

vis[i]=0;

}

sum=0;

}

int main()

{

int i,j,k,sum,flag1,flag2;

while(scanf("%d%d%d%d",&n,&l,&r,&x)!=EOF)

{

t=0;

sum=0;

for(i=1;i<=n;i++)

scanf("%d",&c[i]);

sort(c+1,c+n+1);

if(n==1){

printf("%d\n",t);continue;

}

if(n==2){

sum=c[1]+c[2];

if(sum>=l&&sum<=r&&c[2]-c[1]>=x)printf("1\n");

else printf("0\n");

continue;

}

for(k=2;k<=n;k++)

{

memset(vis,0,sizeof(vis));

memset(p,0,sizeof(p));

m=0;

dfs(1,k,0,0,0);

t=t+m;

}

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

}

return 0;

}