2013
11-26

# Candy Sharing Game

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.

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.

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.

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

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.

#include<cstdio>
using namespace std;
int main()
{
int a[100],n,k,p,t,i,j;
while(scanf("%d",&n),n)
{
k=0;
for(i=0;i<n;i++)  //接受数据
scanf("%d",&a[i]);
while(1)
{
k++;
t=a[n-1]/2;    //t为本轮 a[i]要加多少
for(i=0;i<n;i++)  //循环
{
p=a[i]/2;
a[i]=a[i]/2+t;
t=p;
if(a[i]&1)  //为奇数加1
++a[i];
}
t=a[0];
for(i=0;t==a[i]&&i<n;i++);  //判断是否全部相等
if(i==n)
break;
}
printf("%d %d\n",k,t);
}
return 0;
}

1. Gucci New Fall Arrivals

This is really nice to know. I hope it will be successful in the future. Good job on this and keep up the good work.

2. 代码是给出了，但是解析的也太不清晰了吧！如 13 abejkcfghid jkebfghicda
第一步拆分为 三部分 (bejk, cfghi, d) * C(13,3)，为什么要这样拆分，原则是什么？

3. #include <cstdio>
#include <algorithm>

struct LWPair{
int l,w;
};

int main() {
//freopen("input.txt","r",stdin);
const int MAXSIZE=5000, MAXVAL=10000;
LWPair sticks[MAXSIZE];
int store[MAXSIZE];
int ncase, nstick, length,width, tmp, time, i,j;
if(scanf("%d",&ncase)!=1) return -1;
while(ncase– && scanf("%d",&nstick)==1) {
for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
for(time=-1,i=0;i<nstick;++i) {
tmp=sticks .w;
for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
if(j==time) { store[++time]=tmp; }
else { store[j+1]=tmp; }
}
printf("%dn",time+1);
}
return 0;
}