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.
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.
Print the number of ways to choose a suitable problemset for the contest.
3 5 6 1
1 2 3
2
4 40 50 10
10 20 30 25
2
5 25 35 10
10 10 20 10 20
6
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.
题目信息：从n个数种选择至少两个数，这些数的和在l和r之间，并且这些数的最大值-最小值要大于等于x
思路：由于n个数值很少，那么就采用暴力枚举。关与枚举采用了下面的两种方法：dfs 或者 直接状态压缩枚举所有的状况
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<algorithm>
#define N 100005
using namespace std;
int a[20];
int n, l, r, x;
int cnt;
int ans;
void dfs(int i, int cc, int sum, int minn, int maxn){
if(cc == cnt){
if(sum>=l && sum<=r && maxn - minn>=x)
++ans;
return;
}
if(i>=n) return;
dfs(i+1, cc+1, sum+a[i], min(minn, a[i]), max(maxn, a[i]));
dfs(i+1, cc, sum, minn, maxn);
}
int main(){
cin>>n>>l>>r>>x;
for(int i=0; i<n; ++i)
cin>>a[i];
for(int i=2; i<=n; ++i){
cnt = i;
dfs(0, 0, 0, 10000000, -1);
}
cout<<ans<<endl;
return 0;
}#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<algorithm>
#define N 100005
using namespace std;
int a[20];
int n, l, r, x;
int main(){
cin>>n>>l>>r>>x;
for(int i=0; i<n; ++i)
cin>>a[i];
int s = 1<<n;
int cnt = 0;
for(int i=1; i<s; ++i){
int sum = 0, minn = 10000000, maxn = -1;
for(int j=0; j<n; ++j)
if((1<<j)&i){
sum += a[j];
minn = min(minn, a[j]);
maxn = max(maxn, a[j]);
}
if(sum>=l && sum<=r && maxn-minn>=x)
++cnt;
}
cout<<cnt<<endl;
return 0;
}
