2020.10.30--vj个人赛补题

D - D

Vladik is a competitive programmer. This year he is going to win the International Olympiad in Informatics. But it is not as easy as it sounds: the question Vladik face now is to find the cheapest way to get to the olympiad.

Vladik knows n airports. All the airports are located on a straight line. Each airport has unique id from 1 to n, Vladik's house is situated next to the airport with id a, and the place of the olympiad is situated next to the airport with id b. It is possible that Vladik's house and the place of the olympiad are located near the same airport.

To get to the olympiad, Vladik can fly between any pair of airports any number of times, but he has to start his route at the airport a and finish it at the airport b.

Each airport belongs to one of two companies. The cost of flight from the airport i to the airport j is zero if both airports belong to the same company, and |i - j| if they belong to different companies.

Print the minimum cost Vladik has to pay to get to the olympiad.

Input

The first line contains three integers n, a, and b (1 ≤ n ≤ 10⁵, 1 ≤ a, b ≤ n) — the number of airports, the id of the airport from which Vladik starts his route and the id of the airport which he has to reach.

The second line contains a string with length n, which consists only of characters 0 and 1. If the i-th character in this string is 0, then i-th airport belongs to first company, otherwise it belongs to the second.

Output

Print single integer — the minimum cost Vladik has to pay to get to the olympiad.

Examples

Input

4 1 4
1010

Output

Input

5 5 2
10110

Output

Note

In the first example Vladik can fly to the airport 2 at first and pay |1 - 2| = 1 (because the airports belong to different companies), and then fly from the airport 2 to the airport 4for free (because the airports belong to the same company). So the cost of the whole flight is equal to 1. It's impossible to get to the olympiad for free, so the answer is equal to 1.

In the second example Vladik can fly directly from the airport 5 to the airport 2, because they belong to the same company.

题意：Vladik是一位有竞争力的程序员。今年，他将赢得国际信息学奥林匹克奖。但这并不像听起来那么容易：弗拉迪克现在面临的问题是寻找最廉价的到达奥林匹克运动会的方式。Vladik知道n个机场。所有的机场都位于一条直线上。每个机场的ID从1到n是唯一的，Vladik的房子位于ID为a的机场旁边，奥林匹克公园的位置位于ID为b的机场旁边。弗拉迪克的房子和奥林匹克奥林匹克的所在地可能位于同一机场附近。为了到达奥运会，弗拉迪克可以在任何一对机场之间任意飞行多次，但是他必须在机场a处开始其航线并在机场b处结束。每个机场属于两个公司之一。如果两个机场属于同一公司，则从机场i到机场j的飞行成本为零，如果属于不同的公司，则成本为| i-j |。

输入：第一行包含三个整数n，a和b（1≤n≤105，1≤a，b≤n）-机场数量，弗拉迪克开始其航线的机场ID和机场ID他必须达到的目标。第二行包含一个长度为n的字符串，该字符串仅包含字符0和1。如果该字符串中的第i个字符为0，则第i个机场属于第一家公司，否则属于第二家公司。

输出：打印单个整数-弗拉迪克（Vladik）参加奥林匹克运动会必须支付的最低费用。

题解：可以看出结果不是1就是0，当a==b时结果为0，不同时为1

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n,a,b;

    cin>>n>>a>>b;

    string s;

    for(int i=1;i<=n;i++)cin>>s[i];

    if(s[a]==s[b])cout<<"0"<<endl;

    else cout<<"1"<<endl;

}

E - E

CodeForces - 743B

/*/******/

题意：Chloe和Vladik一样，是个很有竞争力的程序员。和弗拉迪克一样，她没有遇到任何困难，但她对奥运会上提出的任务感到困惑。让我们考虑以下生成整数序列的算法。一开始我们有一个由一个等于1的元素组成的序列。然后执行(n - 1)步。在每一步中，我们取上一步得到的序列，把它附加到它自身的末尾，并在中间插入我们以前没有用过的最小正整数。例如，第一步后得到序列[1,2,1]，第二步后得到序列[1,2,1,3,1,2,1]。任务是在得到的序列中找到索引为k的元素的值(元素从1开始编号)，即经过(n - 1)步。请帮助Chloe解决这个问题!输入仅一行包含两个整数n和k (1<= n<= 50, 1<= k<= 2n - 1)，输出打印单个整数，即得到序列中第k个位置的整数。

题解：

第几串数组中间的数中间数的位置

1 1 1 1

2 121 2 2

3 1213121 3 4

n ......... n 2^(n-1)

先用数组将中间数的位置记录下来，因为中间数的后面都是与前面重复的数，所以可以只将有效长度设为2^(n-1)，因为第几串的值就是中间数的值，所以当k=s[n]时直接输出n值，当k>s[n]时，k-=s[n],n--;直至k=s[n]，输出此时的n值

#include<bits/stdc++.h>

using namespace std;

int main()

{

   int n;

   cin>>n;

   long long k,i,j;

   scanf("%lld",&k);

   long long s[10000]={0};

   for(i=1;i<=51;i++)//记下每步后的中间数的位置

   {

       if(i==1)s[i]=1;

       else s[i]=pow(2,i-1);

   }

   while(k!=s[n])

   {

       if(k>s[n])

       {

           k=k-s[n];

       }

       n--;

   }

   cout<<n<<endl;

}

F - F

CodeForces - 743C

题意：弗拉迪克（Vladik）和克洛伊（Chloe）决定确定他们中谁在数学上更好。弗拉迪克声称，对于任何正整数n，他都可以将分数表示为形式上为三个不同的正分数之和。帮助Vladik，即对于给定的n找到三个不同的正整数x，y和z，使。因为如果数字很大，Chloe无法检查Vladik的答案，他要求您打印不超过10^9的数字。如果没有这样的答案，请打印-1。输入单行包含单个整数n（1≤n≤104）。输出如果存在答案，则打印3个不同的数字x，y和z（1≤x，y，z≤109，x≠y，x≠z，y≠z）。否则打印-1。如果有多个答案，请打印其中的任何一个。示例输入3输出2 7 42输入7输出7 8 56

题解：

n 分数 x y z

1 2/1 无

2 2/2 2 3 6

3 2/3 3 4 12

4 2/4 4 5 20

n 2/n n n+1 n*(n+1)

通过举例可以得到，n=1时结果为-1，其余情况下可以另结果为n n+1 n*(n+1)

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n;

    scanf("%d",&n);

    long long sum=0,num=0,h;

    if(n==1)cout<<"-1"<<endl;

    else

    {

        cout<<n<<" "<<n+1<<" "<<n*(n+1)<<endl;

    }

}