HDU 1034 Candy Sharing Game (模拟）

最新推荐文章于 2024-07-30 10:30:00 发布

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最新推荐文章于 2024-07-30 10:30:00 发布

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原文链接：http://www.cnblogs.com/cmmdc/p/6763839.html

本文介绍了一个关于学生分糖果的游戏模拟程序。学生们围成一圈，在老师的指挥下将手中的糖果平均分配给右侧的同学，若最终糖果数量为奇数，则由老师补充一颗糖果。游戏目标是找出所有学生糖果数量一致时所需的轮数及最终每人拥有的糖果数。

题目链接

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

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.

分析：

题目说的是一个分糖果的游戏，n个学生围成一圈，每个人手上有a[i]个糖果，每一轮游戏开始时，学生手中的糖果都是偶数，他们都把自己手中一半的糖果分给他右边的同学。分完后手中糖果数是奇数的，老师会给他一颗糖果。如果所有人手中的糖果数都一样游戏结束。

输出总共进行了多少轮游戏还有最后学生手中的糖果数。

数组模拟整个游戏的过程即可。

代码：

#include<stdio.h>
int main()
{
    int N,i,a[1000];
    while(~scanf("%d",&N)&&N)
    {
        for(i=1; i<=N; i++)
            scanf("%d",&a[i]);
        int op=0;
        while(1)
        {
            for(i=1; i<N; i++)
                if(a[i]!=a[i+1])
                    break;
            if(i>=N)
                break;
            op++;
            int m=a[N]/2;
            for(i=N; i>1; i--)
            {
                a[i]=a[i]/2;
                a[i]=a[i]+a[i-1]/2;
            }
            a[1]=a[1]/2;
            a[1]=a[1]+m;
            for(i=1; i<=N; i++)
                if(a[i]%2==1)
                    a[i]++;
        }
        printf("%d %d\n",op,a[1]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/cmmdc/p/6763839.html