F - Preparing Olympiad
CodeForces - 550B
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.
Example
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.
题意：每一道题都有一个难度系数，做一个题集使题的难度之和在给定范围内，并且最难的和最容易的题的难度之差要大于给定值，求一共有多少种选题方式。
解法：已知给定的题目数量最大值为15很小，用枚举法，对于一道题只有选和不选两种情况，用二进制法存储选择情况。
代码：
#include<iostream>
using namespace std;
int a[15];
int main()
{
int n,l,r,s;
int ans=0;
cin>>n>>l>>r>>s;
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=1;i<=(1<<n)-1;i++){
int sum=0,num=0;
int minn=9999999,maxn=0;
for(int j=0;j<n;j++)
if((1<<j)&i){
num++;
sum+=a[j];
maxn=max(maxn,a[j]);
minn=min(minn,a[j]);
}
if(sum<=r&&sum>=l&&maxn-minn>=s&&num>1)
ans++;
}
cout<<ans<<endl;
return 0;
}
