Candy Sharing Game
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4942 Accepted Submission(s):
3020
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.
Source
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模拟题，代码绝对不难，关键是看懂题意。
注意：让先给下一个同学才能拿上一个同学的糖果。
题意：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。
附上代码：
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5 int a[1005],n;
6 int xx()
7 {
8 for(int i=1; i<n; i++)
9 {
10 if(a[i]!=a[i-1])
11 return 1;
12 }
13 return 0;
14 }
15 int main()
16 {
17 int m,i,j,k,w;
18 while(~scanf("%d",&n)&&n)
19 {
20 for(i=0; i<n; i++)
21 scanf("%d",&a[i]);
22 m=0;
23 while(xx())
24 {
25 m++;
26 w=a[n-1]/2;
27 for(i=n-1; i>0; i--)
28 {
29 a[i]=a[i]/2+a[i-1]/2;
30 if(a[i]%2==1)
31 a[i]++;
32 }
33 a[0]=a[0]/2+w;
34 if(a[0]%2==1)
35 a[0]++;
36 }
37 printf("%d %d\n",m,a[0]);
38 }
39 return 0;
40 }