CodeForces - Hello 2018 B(树的遍历). C(贪心)

点击打开B题链接

B. Christmas Spruce

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to u. A vertex is called a leaf if it doesn't have children and has a parent.

Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.

The definition of a rooted tree can be found here.

Input

The first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000). Each of the next n - 1 lines contains one integer pi (1 ≤ i ≤ n - 1) — the index of the parent of the i + 1-th vertex (1 ≤ pi ≤ i).

Vertex 1 is the root. It's guaranteed that the root has at least 2 children.

Output

Print "Yes" if the tree is a spruce and "No" otherwise.

Examples

input

4
1
1
1

output

Yes

input

7
1
1
1
2
2
2

output

No

input

8
1
1
1
1
3
3
3

output

Yes

Note

The first example:

The second example:

It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:

题目大意：

给出树节点个数和每个节点的根，求这棵树是不是一颗云杉(每一个非叶子节点满足至少有三个叶子节点)。(没看到“至少”WA了一次)

思路：

构造这棵树然后进行宽搜遍历每一个节点看是否满足云杉条件。

代码：

#include<iostream>
#include<vector>
#include<queue>

using namespace std;
const int maxn = 1050;
vector<int >v[maxn];
queue<int> q;
int n;

int bfs()
{
    int tag = 1;
    q.push(1);
    if(v[1].size() < 3) tag = 0;
    while(q.size() && tag)
    {
        int t = q.front();
        q.pop();
        int cnt = 0;
        for(int i = 0; i < v[t].size(); i++)
        {
            if(v[v[t][i]].size() == 0)
            {
                cnt++;
            }
            else if(v[v[t][i]].size() < 3)
            {
                tag = 0;
                break;
            }
            else q.push(v[t][i]);
        }
        if(cnt < 3 && v[t].size() != 0) tag = 0;
    }
    return tag;
}

int main()
{
    cin >> n;
    for(int i = 2; i <= n; i++)
    {
        int t;
        cin >> t;
        v[t].push_back(i);
    }
    int flag = bfs();
    if(flag) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}

点击打开C题链接

C. Party Lemonade

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A New Year party is not a New Year party without lemonade! As usual, you are expecting a lot of guests, and buying lemonade has already become a pleasant necessity.

Your favorite store sells lemonade in bottles of n different volumes at different costs. A single bottle of type i has volume 2i - 1 liters and costs ci roubles. The number of bottles of each type in the store can be considered infinite.

You want to buy at least L liters of lemonade. How many roubles do you have to spend?

Input

The first line contains two integers n and L (1 ≤ n ≤ 30; 1 ≤ L ≤ 109) — the number of types of bottles in the store and the required amount of lemonade in liters, respectively.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 109) — the costs of bottles of different types.

Output

Output a single integer — the smallest number of roubles you have to pay in order to buy at least L liters of lemonade.

Examples

input

4 12
20 30 70 90

output

150

input

4 3
10000 1000 100 10

output

10

input

4 3
10 100 1000 10000

output

30

input

5 787787787
123456789 234567890 345678901 456789012 987654321

output

44981600785557577

Note

In the first example you should buy one 8-liter bottle for 90 roubles and two 2-liter bottles for 30 roubles each. In total you'll get 12 liters of lemonade for just 150 roubles.

In the second example, even though you need only 3 liters, it's cheaper to buy a single 8-liter bottle for 10 roubles.

In the third example it's best to buy three 1-liter bottles for 10 roubles each, getting three liters for 30 roubles.

题目大意：

给出柠檬水的种类数和节日需要的数量及各种类的价格。已知第i种柠檬水为2^(i-1) L，求最小花费。

思路：

因为相邻柠檬水重量有两倍关系，a[i]的最优解为min(a[i], a[i-1]*2)，同时如果a[i+1] < a[i], a[i] = a[i+1];遍历一次保证每一个二进制位最优，然后进行二进制枚举。如果进制位为1，加上当前位，如果二进制位为0，则要比较当前花费和a[i+1]位花费哪个更小。

代码：

#include<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;
ll a[32];
ll n, m, ans;

int main() {
    cin >> n >> m;
    for(int i = 0; i < n; i++) cin >> a[i];
    for(int i = 1; i < n; i++) {
        a[i] = min(a[i], a[i-1] * 2);
        a[i-1] = min(a[i], a[i-1]);
    }
    for(int i = n; i < 31; i++) a[i] = 2 * a[i-1];
    for(int i = 0; i < 30; i++) {
        if(m & (1 << i)) ans += a[i];
        ans = min(ans, a[i+1]);
    }
    cout << ans << endl;
    return 0;
}