URAL 2056 Scholarship 水题

Scholarship
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86686#problem/D

Description

At last the first term at the University came to its finish. Android Vasya has already passed all the exams and wants to know if he gets a scholarship. There is the following practice of giving scholarship to students at the University:
  • if a student has got satisfactory marks, the scholarship is not given,
  • if a student has passed through the examination period with only excellent marks, he gets a personal scholarship,
  • if a student doesn’t get a personal scholarship and his average mark is not less than 4.5, he gets a high scholarship,
  • if a student gets neither high nor personal scholarship and doesn’t have satisfactory marks, he gets a common scholarship.
A satisfactory mark corresponds to value 3, a good mark corresponds to value 4, and an excellent mark corresponds to value 5. An average mark for a student is the average value of all the marks this student got in his exams. Help Vasya find out which scholarship he gets.

Input

The first line contains an integer n that is the number of exams (1 ≤ n ≤ 10). In the i-th of the next n lines there is an integer mi that is value of Vasya’s mark in i-th exam (3 ≤ mi ≤ 5).

Output

If Vasya doesn’t get any scholarship output “None”. If he gets a common scholarship output “Common”, if he gets a high scholarship output “High”, if he gets a personal one output “Named”.

Sample Input

3
5
5
4

Sample Output

High

HINT

 

题意

给n门课,如果有一门为3分,那就没有奖学金,如果所有都是5分,那就单人奖学金,如果平均分超过4.5,就High奖学金,其他就普通奖学金

题解

水题,读完题就A了……

代码:

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 20001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

int a[1000];
int main()
{
    int n=read();
    int flag=0;
    int sum=0;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        if(a[i]==3)
            flag=1;
        sum+=a[i];
    }
    if(flag)
    {
        cout<<"None"<<endl;
    }
    else if(sum==5*n)
    {
        cout<<"Named"<<endl;
    }
    else if(sum>=4.5*(double)n)
    {
        cout<<"High"<<endl;
    }
    else
        cout<<"Common"<<endl;
        
}

 

posted @ 2015-08-10 18:51  qscqesze  阅读(376)  评论(0编辑  收藏  举报