hdoj 1034 Candy Sharing Game 【模拟】

Candy Sharing Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4422    Accepted Submission(s): 2696

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.
 

 

题意：有N个孩子围成一圈，初始每个人都有一些糖果。以后每轮进行这样的操作——所有的孩子同时分成他右边孩子一些糖果，数量为自己糖果的一半。每轮操作后，若某个孩子的糖果数目为奇数，则老师会发给该孩子一个糖果，当所有孩子的糖果数目相同时游戏结束。问进行的操作数和结束时孩子的糖果数。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define INF 0x3f3f3f
#define eps 1e-4
#define MAXN (10000+10)
#define MAXM (100000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int a[MAXN], pre[MAXN];
int judge(int n)
{
    int num;
    if(a[1] & 1) a[1]++;
    num = a[1];
    for(int i = 2; i <= n; i++)
        if(a[i] & 1)
            a[i]++;
    for(int i = 1; i <= n; i++)
        if(num != a[i])
            return -1;
    return num;
}
int main()
{
    int N;
    while( Ri(N), N)
    {
        for(int i = 1; i <= N; i++)
            Ri(a[i]), pre[i] = a[i];
        a[0] = a[N]; pre[0] = pre[N];
        int cnt = 0;
        while(1)
        {
            for(int i = 1; i <= N; i++)
            {
                a[i] -= pre[i] / 2;
                a[i] += pre[i-1] / 2;
            }
            cnt++;
            int flag = judge(N);
            if(flag != -1)
            {
                printf("%d %d\n", cnt, flag);
                break;
            }
            for(int i = 1; i <= N; i++)
                pre[i] = a[i];
            pre[0] = pre[N];
        }
    }
    return 0;
}