Codeforces 851A

链接:

  http://codeforces.com/contest/851/problem/A

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题目：

Arpa is researching the Mexican wave.

There are n spectators in the stadium, labeled from 1 to n. They start the Mexican wave at time 0.

At time 1, the first spectator stands.
At time 2, the second spectator stands.
…
At time k, the k-th spectator stands.
At time k + 1, the (k + 1)-th spectator stands and the first spectator sits.
At time k + 2, the (k + 2)-th spectator stands and the second spectator sits.
…
At time n, the n-th spectator stands and the (n - k)-th spectator sits.
At time n + 1, the (n + 1 - k)-th spectator sits.
…
At time n + k, the n-th spectator sits.
Arpa wants to know how many spectators are standing at time t.

Input

The first line contains three integers n, k, t (1 ≤ n ≤ 109, 1 ≤ k ≤ n, 1 ≤ t < n + k).

Output

Print single integer: how many spectators are standing at time t.

Examples

input
10 5 3
output
3
input
10 5 7
output
5
input
10 5 12
output
3

Note

In the following a sitting spectator is represented as -, a standing spectator is represented as ^.

At t = 0  ———- number of standing spectators = 0.
At t = 1  ^——— number of standing spectators = 1.
At t = 2  ^^——– number of standing spectators = 2.
At t = 3  ^^^——- number of standing spectators = 3.
At t = 4  ^^^^—— number of standing spectators = 4.
At t = 5  ^^^^^—– number of standing spectators = 5.
At t = 6  -^^^^^—- number of standing spectators = 5.
At t = 7  –^^^^^— number of standing spectators = 5.
At t = 8  —^^^^^– number of standing spectators = 5.
At t = 9  —-^^^^^- number of standing spectators = 5.
At t = 10 —–^^^^^ number of standing spectators = 5.
At t = 11 ——^^^^ number of standing spectators = 4.
At t = 12 ——-^^^ number of standing spectators = 3.
At t = 13 ——–^^ number of standing spectators = 2.
At t = 14 ———^ number of standing spectators = 1.
At t = 15 ———- number of standing spectators = 0.

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题意：

  有n个人，最多有k个人站起来，每秒都有一个人站起来或者是坐下，详情见题目里的note。

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思路：

  t小于k的时候直接输出t，t大于n的时候就是k−(t−n)。

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实现：

#include <bits/stdc++.h>using namespace std;int main() {#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif    long long n, k, t;    scanf("%lld%lld%lld", &n, &k, &t);    if(t<k) return 0*printf("%lld\n", t);    if(t>n) return 0*printf("%lld\n", k-t+n);    return 0*printf("%lld\n", k);}