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  • CodeForces 651B Beautiful Paintings 贪心

    A. Joysticks
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at a1 percent and second one is charged at a2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).

    Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.

    Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.

    Input

    The first line of the input contains two positive integers a1 and a2 (1?≤?a1,?a2?≤?100), the initial charge level of first and second joystick respectively.

    Output

    Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.

    Examples
    input
    3 5
    
    output
    6
    
    input
    4 4
    
    output
    5
    

    题意:在一个序列中,当遇到递增的数时,会觉得开心,现在给一组数,问可以开心几次
    题解:先将序列从小到大排序,记录所有重复出现的数的次数,去掉出现次数最多的求剩下的数的和即可

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int n,a[1100],b[1100];
    bool cmp(int x,int y)
    {
        return x>y;
    }
    int main()
    {
        int i,j,ans;
        while(scanf("%d",&n)!=EOF)
        {
            ans=0;
            memset(b,0,sizeof(b));
            for(i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
                b[a[i]]++;
            }
            sort(b,b+1100,cmp);
            for(i=1;i<1100;i++)
                ans+=b[i];
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Noevon/p/5664812.html
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