time limit per test 2 seconds
memory limit per test 256 megabytes
input standard input
output standard output
This year, as in previous years, MemSQL is inviting the top 25 competitors from the Start[c]up qualification round to compete onsite for the final round. Not everyone who is eligible to compete onsite can afford to travel to the office, though. Initially the top 25 contestants are invited to come onsite. Each eligible contestant must either accept or decline the invitation. Whenever a contestant declines, the highest ranked contestant not yet invited is invited to take the place of the one that declined. This continues until 25 contestants have accepted invitations.
After the qualifying round completes, you know K of the onsite finalists, as well as their qualifying ranks (which start at 1, there are no ties). Determine the minimum possible number of contestants that declined the invitation to compete onsite in the final round.
Input
The first line of input contains K (1 ≤ K ≤ 25), the number of onsite finalists you know. The second line of input contains r1, r2, ..., rK(1 ≤ ri ≤ 106), the qualifying ranks of the finalists you know. All these ranks are distinct.
Output
Print the minimum possible number of contestants that declined the invitation to compete onsite.
Examples
input
25
2 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 22 23 24 25 26 28
output
3
input
5
16 23 8 15 4
output
0
input
3
14 15 92
output
67
Note
In the first example, you know all 25 onsite finalists. The contestants who ranked 1-st, 13-th, and 27-th must have declined, so the answer is 3.
【翻译】一场比赛本来只选择排名前25位的人进入决赛。但是这些人可以有人弃权，如果一个人弃权，那么就在剩下没有入围的人中选出最高分并邀请他参加决赛。当然这个人也可以拒绝，那么就要一直重复上述过程。给出n个人的排名(1<=n<=25),且这n个人都进入决赛了，请求出满足这些人进决赛,最少有多少个人拒绝参加决赛。
题解：
①题意很难懂。
②转化后问题：给出n个数，求最大值，如果最大值大于25输出最大值-25的值，否则输出0。
#include<stdio.h>
using namespace std;
int n,a,b;
int main()
{
scanf("%d",&n);
while(n--)scanf("%d",&b),a=a<b?b:a;
printf("%d",a<25?0:a-25);return 0;
}//Paul_Guderian