一个人可以站立的不同位置的数量
一个人站在 n 个人的队伍中，但他不知道自己到底占据了哪个位置。他可以说，站在他面前的不下于‘f’人，站在他身后的不下于‘b’人。任务是找到他能占据的不同位置的数量。
示例:
Input: n = 3, f = 1, b = 1
Output: 2
3 is the number of people in the line and there can be no less than 1 people standing
in front of him and no more than 1 people standing behind him.So the positions could be 2 and 3
(if we number the positions starting with 1).
Input: n = 5, f = 2, b = 3
Output: 3
In this example the positions are 3, 4, 5\.
进场:我们对每一项进行迭代，检查是否适合条件 a < =i-1 和 n-i < =b(对于 I 从 1 到 n)。第一个条件可以转换成 a+1 < =i ，条件 n-i < =b 在 n-b <中=i ，那么一般条件可以写成 max(a+1，n-b) < =i ，然后我们的答案就可以用公式 n-max(a+1，n-b)+1 来计算了。 以下是上述方法的实施:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the position
int findPosition(int n, int f, int b)
{
return n - max(f + 1, n - b) + 1;
}
// Driver code
int main()
{
int n = 5, f = 2, b = 3;
cout << findPosition(n, f, b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// Function to find the position
static int findPosition(int n, int f, int b)
{
return n - Math.max(f + 1, n - b) + 1;
}
// Driver code
public static void main(String args[])
{
int n = 5, f = 2, b = 3;
System.out.print(findPosition(n, f, b));
}
}
Python 3
# Python3 implementation of
# above approach
# Function to find the position
def findPosition(n, f, b):
return n - max(f + 1, n - b) + 1;
# Driver code
n, f, b = 5, 2, 3
print(findPosition(n, f, b))
# This code is contributed by
# Sanjit_Prasad
C
// C# implementation of above approach
using System;
class GFG
{
// Function to find the position
static int findPosition(int n,
int f, int b)
{
return n - Math.Max(f + 1, n - b) + 1;
}
// Driver code
public static void Main()
{
int n = 5, f = 2, b = 3;
Console.WriteLine(findPosition(n, f, b));
}
}
// This code is contributed
// by inder_verma
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of above approach
// Function to find the position
function findPosition($n, $f, $b)
{
return $n - max($f + 1,
$n - $b) + 1;
}
// Driver code
$n = 5; $f = 2; $b = 3;
echo findPosition($n, $f, $b);
// This code is contributed
// by anuj_67
?>
java 描述语言
<script>
// Java Script implementation of above approach
// Function to find the position
function findPosition(n,f,b)
{
return n - Math.max(f + 1, n - b) + 1;
}
// Driver code
let n = 5, f = 2, b = 3;
document.write(findPosition(n, f, b));
//contributed by 171fa07058
</script>
Output:
3
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