Codeforces Round #306 (Div. 2), problem: (B) Preparing Olympiad【dfs或01枚举】

 

 

题意： 给出n个数字，要求在这n个数中选出至少两个数字，使得它们的和在l,r之间，并且最大的与最小的差值要不小于x。n<=15

Problem - 550B - Codeforces

二进制

利用二进制, 第i位为1则加上a[i], 为0则不加, 

#include<iostream>
#include <algorithm>
#include <cmath>
#include<map>

using namespace std;
typedef long long LL;
int a[20];
int main()
{
    int n,l,r,x, res=0;
    cin >> n >> l >> r >>x;
    for(int i = 0; i < n; i ++)cin >> a[i];
    for(int i =1; i < (1<<n); i ++)
    {
        int idx = 0, sum = 0;
        int minn = 999999999, maxx = 0;
        for(int j = i; j; j >>=1, idx ++)
            if(j&1==1)
            {
                sum += a[idx];
                maxx = max(maxx,a[idx]);
                minn = min(minn, a[idx]);
            }
                
        if(maxx -minn >= x && sum <= r && sum >= l)
            res ++;
    }
    
    cout << res <<'\n';
    
    return 0;
}

dfs方法

#include<iostream>
#include <algorithm>
#include <cmath>
#include<map>

using namespace std;
typedef long long LL;
int a[20];
int n,l,r,x, res=0;

void dfs(int sum, int maxx, int minn, int num)
{
    if(maxx - minn >= x && sum >= l && sum <= r && num==n)
        res ++;
        
    if(num==n)return;
    dfs(sum + a[num], max(maxx, a[num]), min(minn, a[num]), num +1);
    
    dfs(sum, maxx, minn, num+1);
    
}
int main()
{
    
    cin >> n >> l >> r >>x;
    for(int i = 0; i < n; i ++)cin >> a[i];
    
    dfs(0,0,999999999,0);
    
    cout << res <<'\n';
    
    return 0;
}