杭电 HDU 1034 Candy Sharing Game

Candy Sharing Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3516    Accepted Submission(s): 2190

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

6362222211222018161412108642424680

 

Sample Output

15 1417 224 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

Greater New York 2003

 

水~   题意是哨声 响起 围成一圈的每个学生 都拿出自己偶数糖块的部分，同时的给右边的人，一次下来 如果某个同学 剩余奇数块儿 那么老师会补充一块，知道判断所有

学生糖块数目相同时 终止  输出游戏次数 和最后每个同学相等时的糖块儿数。

#include<iostream>using namespace std;int main(){int ls[100],count,i,N,t;while(cin>>N,N){ count=0;for(i=1;i<=N;i++)cin>>ls[i];for(int k=0;;count++,k++){    int y=ls[N]=ls[N]/2;for(int j=N;j>1;j--){    ls[j-1]/=2;ls[j]+=ls[j-1];}ls[1]+=y;for(int m=1;m<=N;m++)if(ls[m]%2==1)ls[m]+=1;    for( t=2;t<=N;t++){if(ls[t]!=ls[t-1]){break;}}if(t>N){cout<<count+1<<" "<<ls[t-1]<<endl;break;}elsecontinue;}}return 0;}