CodeCraft-19 and Codeforces Round #537 (Div. 2) C. Creative Snap 分治

CodeCraft-19 and Codeforces Round #537 (Div. 2) C. Creative Snap 分治

Thanos wants to destroy the avengers base, but he needs to destroy the avengers along with their base.

Let we represent their base with an array, where each position can be occupied by many avengers, but one avenger can occupy only one position. Length of their base is a perfect power of 2

. Thanos wants to destroy the base using minimum power. He starts with the whole base and in one step he can do either of following:

  • if the current length is at least 2

, divide the base into 2

 

  • equal halves and destroy them separately, or
  • burn the current base. If it contains no avenger in it, it takes A

amount of power, otherwise it takes his Bnal amount of power, where na is the number of avengers and l

  • is the length of the current base.
Output the minimum power needed by Thanos to destroy the avengers' base.
Input

The first line contains four integers n

, k, A and B (1n30, 1k105, 1A,B104), where 2n is the length of the base, k is the number of avengers and A and B

are the constants explained in the question.

The second line contains k

integers a1,a2,a3,,ak (1ai2n), where ai

represents the position of avenger in the base.

Output

Output one integer — the minimum power needed to destroy the avengers base.

Examples
Input
Copy
2 2 1 2
1 3
Output
Copy
6
Input
Copy
3 2 1 2
1 7
Output
Copy
8
Note

Consider the first example.

One option for Thanos is to burn the whole base 14

with power 224=16

.

Otherwise he can divide the base into two parts 12

and 34

.

For base 12

, he can either burn it with power 212=4 or divide it into 2 parts 11 and 22

.

For base 11

, he can burn it with power 211=2. For 22, he can destroy it with power 1, as there are no avengers. So, the total power for destroying 12 is 2+1=3, which is less than 4

.

Similarly, he needs 3

power to destroy 34. The total minimum power needed is 6.

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/
int n, k;
int A, B;
int a[maxn];

ll dfs(ll l, ll r) {
	ll L = lower_bound(a + 1, a + 1 + k, l) - a;
	ll R = upper_bound(a + 1, a + 1 + k, r) - a; R--;
	ll tot = R - L + 1;
	ll cost = 0;
	if (tot == 0)cost = A;
	else cost = B * (r - l + 1)*tot;
	if (l == r || tot == 0)return cost;
	ll mid = (l + r) >> 1;
	return min(cost, dfs(l, mid) + dfs(mid+1, r));
}

int main()
{
	//	ios::sync_with_stdio(0);
	rdint(n); rdint(k); rdint(A); rdint(B);
	for (int i = 1; i <= k; i++)rdint(a[i]);
	sort(a + 1, a + 1 + k);
	ll x = (ll)(1 << n);
	cout << (ll)dfs(1, x) << endl;
	return 0;
}

 

posted @ 2019-02-10 15:24 NKDEWSM 阅读( ...) 评论( ...) 编辑 收藏

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