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  • Codeforces gym 100685 C. Cinderella 水题

    C. Cinderella
    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://codeforces.com/gym/100685/problem/C

    Description

    Cinderella is given a task by her Stepmother before she is allowed to go to the Ball. There are N (1 ≤ N ≤ 1000) bottles with water in the kitchen. Each bottle contains Li (0 ≤ Li ≤ 106) ounces of water and the maximum capacity of each is 109 ounces. To complete the task Cinderella has to pour the water between the bottles to fill them at equal measure.

    Cinderella asks Fairy godmother to help her. At each turn Cinderella points out one of the bottles. This is the source bottle. Then she selects any number of other bottles and for each bottle specifies the amount of water to be poured from the source bottle to it. Then Fairy godmother performs the transfusion instantly.

    Please calculate how many turns Cinderella needs to complete the Stepmother's task.

    Input

    The first line of input contains an integer number N (1 ≤ N ≤ 1000) — the total number of bottles.

    On the next line integer numbers Li are contained (0 ≤ Li ≤ 106) — the initial amount of water contained in ith bottle.

    Output

    Output a single line with an integer S — the minimal number of turns Cinderella needs to complete her task.

    Sample Input

    3
    5 7 7

    Sample Output

    2

    HINT

    题意

    每一个回合可以选择一瓶水然后给其他杯子倒水,然后问你最少几个回合可以使得所有的杯子里面的水都一样

    题解

    求一个平均数,然后大于平均数的就直接倒水就好了

    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 2000001
    #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;
    }
    //**************************************************************************************
    
    ll sum=0;
    ll a[maxn];
    int main()
    {
        int n=read();
        for(int i=0;i<n;i++)
            a[i]=read(),sum+=a[i];
        ll kiss=sum/n;
        int ans=0;
        for(int i=0;i<n;i++)
            if(a[i]>kiss)
                ans++;
        cout<<ans<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4702807.html
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