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.
題意：每一道題都有一個難度系數(shù)低千，做一個題集使題的難度之和在給定范圍內(nèi)，并且最難的和最容易的題的難度之差要大于給定值，求一共有多少種選題方式申窘。
解法：已知給定的題目數(shù)量最大值為15很小颁股，用枚舉法肝谭，對于一道題只有選和不選兩種情況谴返，用二進(jìn)制法存儲選擇情況结榄。
代碼：
#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;
}
