2020.4.12--个人赛

A - Balloons

There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.

Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$

They want to divide the balloons among themselves. In addition, there are several conditions to hold:

Do not rip the packets (both Grigory and Andrew should get unbroken packets);
Distribute all packets (every packet should be given to someone);
Give both Grigory and Andrew at least one packet;
To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.

Help them to divide the balloons or determine that it's impossible under these conditions.

Input

The first line of input contains a single integer $n$

The second line contains $n$

Output

If it's impossible to divide the balloons satisfying the conditions above, print $- 1$

Otherwise, print an integer $k$

If there are multiple ways to divide balloons, output any of them.

Examples

Input

3
1 2 1

Output

2
1 2

Input

2
5 5

Output

-1

Input

1
10

Output

-1

Note

In the first test Grigory gets $3$

In the second test there's only one way to divide the packets which leads to equal numbers of balloons.

In the third test one of the boys won't get a packet at all.

题意：两个人买了n袋气球，每袋气球里面有ai个气球，现在要把这些气球分给两个人，满足下列要求：

不要撕毁包(格里戈里和安德鲁都应该得到完整的包)；
分发所有的数据包(每个包都应该给某人)；
给格里戈里和安德鲁至少一包；
为了提供更多的乐趣，Grigory包中的气球总数不应该等于Andrew包中的气球总数。
暴力把总发气球数/2，注意精度，用浮点数，取整会错，如果有一袋气球不等于sum/2，就输出，否则就没有这样的结果

#include<bits/stdc++.h>
using namespace std;
map<int,int>m;
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int n,a[100],sum=0,flag=0;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>a[i],sum+=a[i],m[a[i]]=i+1;
    sort(a,a+n,cmp);
    if(n>1)
    {
        for(int i=0; i<n; i++)
            if (a[i]<sum*1.0/2)
            {
                cout<<1<<endl<<m[a[i]]<<endl;
                flag=1;
                break;
            }
    }
if(!flag)
    cout<<-1<<endl;
    return 0;

}

B - Cutting

There are a lot of things which could be cut — trees, paper, "the rope". In this problem you are going to cut a sequence of integers.

There is a sequence of integers, which contains the equal number of even and odd numbers. Given a limited budget, you need to make maximum possible number of cuts such that each resulting segment will have the same number of odd and even integers.

Cuts separate a sequence to continuous (contiguous) segments. You may think about each cut as a break between two adjacent elements in a sequence. So after cutting each element belongs to exactly one segment. Say, $[4, 1, 2, 3, 4, 5, 4, 4, 5, 5]$

The cost of the cut between $x$

Input

First line of the input contains an integer $n$

Second line contains $n$

Output

Print the maximum possible number of cuts which can be made while spending no more than $B$

Examples

Input

6 4
1 2 5 10 15 20

Output

Input

4 10
1 3 2 4

Output

Input

6 100
1 2 3 4 5 6

Output

Note

In the first sample the optimal answer is to split sequence between $2$

In the second sample it is not possible to make even one cut even with unlimited number of bitcoins.

In the third sample the sequence should be cut between $2$

题意：给出一段数列，里面有N个数ai，奇数个数和偶数个数相同，然后你可以在奇数和偶数个数相等的时候，你就可以剪一个片段，需要花费|a[i]-a[i+1]|。现在总共预算是b，让你在b预算内，尽可能的多剪片段，输出最多的剪的次数。

题解：贪心先把所有能减的片段，奇数和偶数个数相等的时候，把花费存起来，然后排个序，从前面累加，如果没有超出预算b剪的次数就++

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m,a[110],ans=0,cnt1=0,cnt2=0,b[110],c=0,sum=0;
    cin>>n>>m;
    for(int i=0;i<n;i++)
         cin>>a[i];
    for(int i=0;i<n;i++)
    {
        if(a[i]%2) cnt1++;
         else cnt2++;
         if(cnt1==cnt2&&i+1<n)
           b[c++]=abs(a[i]-a[i+1]);
    }
    sort(b,b+c);
    for(int i=0;i<c;i++)
    {
        sum+=b[i];
        if(sum<=m)
            ans++;
        else break;
    }
    cout<<ans;
    return 0;
}

C - Convert to Ones

You've got a string $a_{1}, a_{2}, \dots, a_{n}$

Let's call a sequence of consecutive elements $a_{i}, a_{i + 1}, \dots, a_{j}$

You can apply the following operations any number of times:

Choose some substring of string $a$
Choose some substring of string $a$

You can apply these operations in any order. It is allowed to apply the operations multiple times to the same substring.

What is the minimum number of coins you need to spend to get a string consisting only of ones?

Input

The first line of input contains integers $n$

The second line contains the string $a$

Output

Print a single integer — the minimum total cost of operations you need to spend to get a string consisting only of ones. Print $0$

Examples

Input

5 1 10
01000

Output

Input

5 10 1
01000

Output

Input

7 2 3
1111111

Output

Note

In the first sample, at first you need to reverse substring $[1 \dots 2]$

Then the string was changed as follows:

«01000» $\to$

The total cost of operations is $1 + 10 = 11$

In the second sample, at first you need to invert substring $[1 \dots 1]$

Then the string was changed as follows:

«01000» $\to$

The overall cost is $1 + 1 = 2$

In the third example, string already consists only of ones, so the answer is $0$

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAXN = 300005;
char s[MAXN];
int main() {
    int n, a, b;
    while(scanf("%d %d %d", &n, &a, &b) != EOF) {
        scanf("%s", s);
        long long temp = 0;
        int len = strlen(s);
        for(int i = 0; i < len; i++) {
            int flag = i;
            if(s[flag] == '0') {
                temp++;
                while(s[flag] == '0' && flag < len)
                    flag++;
                i = flag;
            }
        }
        if(temp == 0) {
            printf("0\n");
        } else {
            temp--;
            printf("%lld\n", 1ll*min(temp*a + b, b*(temp + 1)));
        }
    }
    
}

D - Sonya and Hotels

Sonya decided that having her own hotel business is the best way of earning money because she can profit and rest wherever she wants.

The country where Sonya lives is an endless line. There is a city in each integer coordinate on this line. She has $n$

Sonya understands that her business needs to be expanded by opening new hotels, so she decides to build one more. She wants to make the minimum distance from this hotel to all others to be equal to $d$

Because Sonya is lounging in a jacuzzi in one of her hotels, she is asking you to find the number of cities where she can build a new hotel so that the minimum distance from the original $n$

Input

The first line contains two integers $n$

The second line contains $n$

Output

Print the number of cities where Sonya can build a new hotel so that the minimum distance from this hotel to all others is equal to $d$

Examples

Input

4 3
-3 2 9 16

Output

Input

5 2
4 8 11 18 19

Output

Note

In the first example, there are $6$

In the second example, there are $5$

题意：

在一条直线上有n家宾馆，现在要新建一家宾馆，要求新建的宾馆与原来的n家宾馆的最小距离是d，且每个位置只能建一家宾馆

求满足要求要求的位置的个数

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
const int N = 105;
map<int,int>mp;
int arr[N];
int check(int n,int x,int d){
    for(int i=0;i<n;i++){
        if(fabs(arr[i]-x)<d) return 0;
    }
    return 1;
}
int main(){
    int n,d;
    scanf("%d%d",&n,&d);
    for(int i=0;i<n;i++){
        scanf("%d",&arr[i]);
        mp[arr[i]]=1;
    }  
    int ans=0;
    for(int i=0;i<n;i++){
        int x=arr[i]-d;
        int y=arr[i]+d;
        if(check(n,x,d)&&!mp[x]){
          ans++;
          mp[x]=1;    
        }
        if(check(n,y,d)&&!mp[y]){
          ans++; 
          mp[y]=1;
        }
    }
    printf("%d\n",ans);
    }

E - Sonya and Exhibition

Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition.

There are $n$

She knows that exactly $m$

Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.

Input

The first line contains two integers $n$

Each of the next $m$

Output

Print the string of $n$

If there are multiple answers, print any.

Examples

Input

Output

Input

Output

Note

In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions;

in the segment $[1 \dots 3]$
in the segment $[2 \dots 4]$
in the segment $[2 \dots 5]$

The total beauty is equal to $2 + 2 + 4 = 8$

In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions;

in the segment $[5 \dots 6]$
in the segment $[1 \dots 4]$
in the segment $[4 \dots 6]$

The total beauty is equal to $1 + 4 + 2 = 7$

题意：小明家有一个长为n的花园（只有玫瑰花和百合花，且花只能放在整数位置，每个位置放一朵），有m个小朋友来他家参观，每个小朋友将会以li为起点，参观到ri。小朋友获得的愉悦感等同与从l到r的距离里，玫瑰花与百合花数量的乘积。问小明怎样摆放花才能使所以的小朋友获得的愉悦感最大，分别用01表示玫瑰花和百合花。

思路：高中学过，a+b=t，要使a×b最大，那么，a和b应该尽量接近。也就是说，如果a+b=3，那么a=1，b=2，或者b=1，a=2。如果a+b=4，那么a=2，b=2。这样的话，不管小朋友怎么参观，我们只要将玫瑰和百合间隔种植就可以了

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++) cout<<(i&1);
    cout<<endl;
   }

F - Sonya and Robots

Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers.

Sonya has drawn $n$

Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one.

For example, if the numbers $[1, 5, 4, 1, 3]$

Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot.

Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ( $p$

Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet.

Input

The first line contains a single integer $n$

The second line contains $n$

Output

Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet.

Examples

Input

5
1 5 4 1 3

Output

Input

7
1 2 1 1 1 3 2

Output

Note

In the first example, Sonya can give pairs ( $1$

In the second example, Sonya can give pairs ( $1$

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
deque <char> q;
map<ll,ll> m;
set<ll> s;
int main()
{
    ll n,i,a[101001],b[100101];
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    for(i=n-1;i>=0;i--)
    {
        b[i]=s.size();
        s.insert(a[i]);
    }
    ll ans=0;
    for(i=0;i<n;i++)
    {
        if(m[a[i]]==0)
            ans+=b[i];
        m[a[i]]++;
    }
    cout<<ans<<endl;
  }

A - Balloons

B - Cutting

C - Convert to Ones

D - Sonya and Hotels

F - Sonya and Robots

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