Codeforces Beta Round #92 (Div. 2) / 124A The number of positions (简单数学)

原创于 2013-08-22 22:22:59 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces #acm

acm之路--数学同时被 2 个专栏收录

262 篇文章

订阅专栏

Codeforces

83 篇文章

订阅专栏

本文介绍了一个编程问题：已知Petr在线中且至少有a人在其前，最多有b人在其后，求Petr可能的位置数量。输入包括人数n及a和b的值，输出则是所有可能位置的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Petr stands in line of n people, but he doesn't know exactly which position he occupies. He can say that there are no less than a people standing in front of him and no more than b people standing behind him. Find the number of different positions Petr can occupy.

Input

The only line contains three integers n, a and b (0 ≤ a, b < n ≤ 100).

Output

Print the single number — the number of the sought positions.

   Sample test(s)
 
     input
   
3 1 1

     output
   
2

     input
   
5 2 3

     output
   
3

Note

The possible positions in the first sample are: 2 and 3 (if we number the positions starting with 1).

In the second sample they are 3, 4 and 5.

/*30ms,0KB*/

#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
	int n, a, b;
	scanf("%d%d%d", &n, &a, &b);
	printf("%d", n - max(a + 1, n - b) + 1);
	return 0;
}