文章标题 POJ :1666 Candy Sharing Game（模拟）

Candy Sharing Game

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

Notes:
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
Source

Greater New York 2003

题意：有n个学生，围成一个圈，刚开始每个人都有偶数个糖，然后每次每个人把自己的一半分给右边的人，如果拿了旁边的糖后总数为奇数个，那老师会再给你一个糖变成偶数，然后要我们求总共用几次所有人的糖数目相同。
分析：直接模拟，把每个人的糖数目放进数组，然后就每次减少一半和加上旁边人给的，最后判断是否所有人的糖数目相同。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int n; 
void time(int a[],int n){
    int b[n];//将a数组复制到b数组
    for (int i=0;i<n;i++) b[i]=a[i]; 
    for (int i=1;i<n;i++){
        if ((b[i]+b[i-1])%2==0){
            a[i]=(b[i]+b[i-1])/2;
        }
        else a[i]=(b[i]+b[i-1])/2+1;
    }
    if ((b[0]+b[n-1])%2==0){//最后一个和第一个
        a[0]=(b[0]+b[n-1])/2;
    } 
    else a[0]=(b[0]+b[n-1])/2+1;
}
bool judge (int a[],int n){//判断是否每个人的糖数目相同
    for (int i=0;i<n-1;i++){
        if (a[i]!=a[i+1])return false;//有一个不相同直接return
    }
    if (a[0]!=a[n-1])return false;
    return true;
}
int main ()
{
    while (scanf ("%d",&n)&&n){
        int a[n];
        for(int i=0;i<n;i++){
            scanf ("%d",&a[i]);
            a[i]/=2;//先减半
        } 
        int cnt=0;
        while (!judge(a,n)){
            cnt++;
            time(a,n);
        }
        printf ("%d %d\n",cnt,a[0]*2);//最后得乘以2
    }       
    return 0;
}