CodeForces 570B Simple Game 概率

原题： http://codeforces.com/contest/570/problem/B

题目：

Simple Game

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

One day Misha and Andrew were playing a very simple game. First, each player chooses an integer in the range from 1 to n. Let’s assume that Misha chose number m, and Andrew chose number a.

Then, by using a random generator they choose a random integer c in the range between 1 and n (any integer from 1 to n is chosen with the same probability), after which the winner is the player, whose number was closer to c. The boys agreed that if m and a are located on the same distance from c, Misha wins.

Andrew wants to win very much, so he asks you to help him. You know the number selected by Misha, and number n. You need to determine which value of a Andrew must choose, so that the probability of his victory is the highest possible.

More formally, you need to find such integer a (1 ≤ a ≤ n), that the probability that is maximal, where c is the equiprobably chosen integer from 1 to n (inclusive).

Input

The first line contains two integers n and m (1 ≤ m ≤ n ≤ 109) — the range of numbers in the game, and the number selected by Misha respectively.

Output

Print a single number — such value a, that probability that Andrew wins is the highest. If there are multiple such values, print the minimum of them.

Sample test(s)

input

3 1

output

2

input

4 3

output

2

Note

In the first sample test: Andrew wins if c is equal to 2 or 3. The probability that Andrew wins is 2 / 3. If Andrew chooses a = 3, the probability of winning will be 1 / 3. If a = 1, the probability of winning is 0.

In the second sample test: Andrew wins if c is equal to 1 and 2. The probability that Andrew wins is 1 / 2. For other choices of a the probability of winning is less.

思路：

知道总共能够选的数，和misha选的数。求andrew选一个能让尽可能多的数到andrew的距离比misha更近。

我们再数轴上来看：



显然，假设m到n的长度比1到m大非常多，andrew选择m+1至少右边全部的他都能比misha近。

同理，假设左边比右边长，那么就选m-1这个点。

假设左边和右边相等，题目要求说If there are multiple such values, print the minimum of them.要求我们打印小的。也就是m-1。

特别的。当n=1的时候，仅仅能选择数1。所以输出1。

代码：

#include<iostream>
#include<stdio.h>

using namespace std;
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==1)
            printf("1\n");
        else if(n-m>m-1)
            printf("%d\n",m+1);
        else
            printf("%d\n",m-1);
    }
    return 0;
}