HDU 1538-A Puzzle for Pirates

A bunch of pirates have gotten their hands on a hoard of gold pieces and wish to divide the loot. They are democratic pirates in their own way, and it is their custom to make such divisions in the following manner: The fiercest pirate makes a proposal about the division, and everybody votes on it, including the proposer. If 50 percent or more are in favor, the proposal passes and is implemented forthwith. Otherwise the proposer is thrown overboard, and the procedure is repeated with the next fiercest pirate. 
All the pirates enjoy throwing one of their fellows overboard, but if given a choice they prefer cold, hard cash, the more the better. They dislike being thrown overboard themselves. All pirates are rational and know that the other pirates are also rational. Moreover, no two pirates are equally fierce, so there is a precise pecking order — and it is known to them all. The gold pieces are indivisible, and arrangements to share pieces are not permitted, because no pirate trusts his fellows to stick to such an arrangement. It's every man for himself. Another thing about pirates is that they are realistic. They believe 'a bird in the hand is worth two in the bush' which means they prefer something that is certain than take a risk to get more, where they might lose everything. 

For convenience, number the pirates in order of meekness, so that the least fierce is number 1, the next least fierce number 2 and so on. The fiercest pirate thus gets the biggest number, and proposals proceed in the order from the biggest to the least. 

The secret to analyzing all such games of strategy is to work backward from the end. The place to start is the point at which the game gets down to just two pirates, P1 and P2. Then add in pirate P3, P4, ... , one by one. The illustration shows the results when 3, 4 or 5 pirates try to divide 100 pieces of gold. 



Your task is to predict how many gold pieces a given pirate will get.

The input consists of a line specifying the number of testcases, followed by one line per case with 3 integer numbers n, m, p. n (1 ≤ n ≤ 10^4) is the number of pirates. m (1 ≤ m ≤ 10^7) is the number of gold pieces. p (1 ≤ p ≤ n) indicates a pirate where p = n indicates the fiercest one. 

The output for each case consists of a single integer which is the minimal number of gold pieces pirate p can get. For example, if pirate p can get 0 or 1 gold pieces, output '0'. If pirate p will be thrown overboard, output 'Thrown'. 

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;int main()
{int sum;///存所求人所得金币数目int n,m,p;int t;scanf("%d",&t);while(t--){sum=-1;///初始scanf("%d %d %d",&n,&m,&p);if(n<=2*m+1)///说明金币够分的情况{if(p==n)///如果正好是求最强大的人所得金币数sum=m-(n-1)/2;else if((p%2)==(n%2))///如果所求人的奇偶性与n相同，则都能得到一个金币sum=1;else///否则得话，就没有sum=0;}else{int bj=0;int a;for(int i=20;i>=1;i--){if(n==2*m+(1<<i))/**如果正好是2*m+2^i,那么是稳定状态，只要将m个金币分出去就可以保命，所以具体是谁可能得到金币，我们不确定，这种情况所有的结构都是0**/{sum=0;bj=1;break;}else if(n>(2*m+(1<<i))&&n<(2*m+(1<<(i+1))))///如果n在两种稳定的情况之间，那么记录下最小的稳定状态{a=2*m+(1<<i);break;}}if(bj==0){if(p>a)///超过稳定状态的都是必死的sum=-1;else///其余的都是稳定状态的sum=0;}}if(sum==-1)printf("Thrown\n");elseprintf("%d\n",sum);}return 0;
}