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  • 水题:HDU1034-Candy Sharing Game

    解题心得:

    1、我是用的模拟的算法直接模拟的每一轮的分配方法的得出的答案,看到有些大神使用的链表做的,好像链表才是这道题的镇长的做法吧。



    题目:
    Candy Sharing Game
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6525    Accepted Submission(s): 3962


    Problem Description
    A number of students sit in a circle facing their teacher in the center. Each student initially has an even number of pieces of candy. When the teacher blows a whistle, each student simultaneously gives half of his or her candy to the neighbor on the right. Any student, who ends up with an odd number of pieces of candy, is given another piece by the teacher. The game ends when all students have the same number of pieces of candy.
    Write a program which determines the number of times the teacher blows the whistle and the final number of pieces of candy for each student from the amount of candy each child starts with.
     

    Input
    The input may describe more than one game. For each game, the input begins with the number N of students, followed by N (even) candy counts for the children counter-clockwise around the circle. The input ends with a student count of 0. Each input number is on a line by itself.
     

    Output
    For each game, output the number of rounds of the game followed by the amount of candy each child ends up with, both on one line.
     

    Sample Input
    6
    36
    2
    2
    2
    2
    2
    11
    22
    20
    18
    16
    14
    12
    10
    8
    6
    4
    2
    4
    2
    4
    6
    8
    0
     

    Sample Output
    15 14
    17 22
    4 8

    Hint

    The game ends in a finite number of steps because:
    1. The maximum candy count can never increase.
    2. The minimum candy count can never decrease.
    3. No one with more than the minimum amount will ever decrease to the minimum.
    4. If the maximum and minimum candy count are not the same, at least one student with the minimum amount must have their count increase.




    #include<bits/stdc++.h>
    using namespace std;
    int a[10000],b[10000];//b用于记录a分配的数目;
    int main()
    {
        int n;
        bool flag;
        while(~scanf("%d",&n))
        {
            if(n == 0)
                break;
            int sum = 0;
            flag = false;
            //初始化输入
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
                if(a[i]%2)
                    a[i]+=1;
            }
            while(!flag)
            {
                sum ++;//记录经过了多少轮;
                for(int i=1;i<=n;i++)
                {
                    b[i] = a[i]/2;
                    a[i]/=2;
                }
                for(int i=2;i<=n;i++)
                    a[i] = a[i] + b[i-1];
                a[1] = a[1] + b[n];
                for(int i=1;i<=n;i++)
                    if(a[i]%2)
                        a[i]++;
                int k;
                for(k=1;k<n;k++)
                    if(a[k] != a[k+1])
                        break;
                if(k == n)
                    flag = true;//用于标记是否已经分配平衡;
            }
            printf("%d %d
    ",sum++,a[n]);
        }
        return 0;
    }
    

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  • 原文地址:https://www.cnblogs.com/GoldenFingers/p/9107346.html
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