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.
The only line contains three integers n, a and b (0 ≤ a, b < n ≤ 100).
Print the single number — the number of the sought positions.
3 1 1
2
5 2 3
3
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;
}

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。
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