CodeForces 570B Simple Game

题目：

Description
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 Input
Input
3 1
Output
2
Input
4 3
Output
2
Hint
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.

题目大意：
两个人分别从1~n中选择一个数，一个选m,另一个选a，m已经确定。
规定一个1~n中的一个随机数c，谁选的数靠近c，谁就胜。如果距离都相同，则选m的人胜。
选a的人要胜利，请输出符合条件的a的最小值。

其实这道题仔细想也不是特别难，如果m在左边，则a就是m+1；如果m在右边，则a就是m-1.如果n和m都为1,则a为1（实际上此时是选m的赢）。

参考代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

int judge(const int n, const int m) {
    if (n == 1 && m == 1) return 1;
    else if (m > n / 2) {
        return m - 1;
    }
    else return m + 1;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int n, m;
    cin >> n >> m;
    int ans = judge(n, m);
    cout << ans << endl;
    return 0;
}