A Puzzle for Pirates

Problem Description

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.

 

Input

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. 

 

Output

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‘. 

 

Sample Input

 

Sample Output

 

Author

Otter

 

Source

TJU Contest August 2006

 

 

本文转自： http://blog.csdn.net/ACM_cxlove?viewmode=contents    

#include<iostream>#include<cstdio>#include<ctime>#include<cstring>#include<cmath>#include<algorithm>#include<cstdlib>#include<vector>#define C    240#define TIME 10#define inf 1<<25#define LL long longusing namespace std;//保存2的幂int fac[15]={2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};void slove(int n,int m,int p){    //金币够贿赂的情况    if(n<=2*m){        //不是决策者，而且奇偶性相同，都能被贿赂        if(n!=p&&(n%2==p%2))            printf("1\n");        //剩下的都是决策者拥有        else if(n==p)            printf("%d\n",m-(n-1)/2);        else        //其余人分不到金币，他们的决策不影响全局            printf("0\n");        return ;    }    //这时候的不同在于决策者不能拿金币    else if(n==2*m+1){        if(p<2*m&&p&1)            printf("1\n");        else            printf("0\n");        return ;    }    int t=n-2*m,i;    //这是刚好保命的情况，对于决策者来说，肯定没有金币    //对于其它人来说，要么肯定没有金币，要么可能没有金币，不确定性    for( i=0;i<14;i++){        if(t==fac[i]){            printf("0\n");            return;        }    }    for( i=1;i<14;i++)        if(t<fac[i]){            //决策者必死            if(p>2*m+fac[i-1]&&p<2*m+fac[i])                 printf("Thrown\n");            else                 printf("0\n");            return ;        }}int main(){    int t,n,m,p;    scanf("%d",&t);    while(t--){        scanf("%d%d%d",&n,&m,&p);        slove(n,m,p);    }    return 0;}