POJ 1704 Georgia and Bob

Description

Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example: 

Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game. 

Georgia always plays first since "Lady first". Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out. 

Given the initial positions of the n chessmen, can you predict who will finally win the game? 

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen.

Output

For each test case, prints a single line, "Georgia will win", if Georgia will win the game; "Bob will win", if Bob will win the game; otherwise 'Not sure'.

Sample Input

2
3
1 2 3
8
1 5 6 7 9 12 14 17

Sample Output

Bob will win
Georgia will win

从恐惧到入门

博弈论!!!对于棋子,我们对他们进行每两个分一组,把每组之间的首和尾之间的空格作为每一堆石子的个数,可以这么想:每组之间的距离对结果造不成影响,因为如果前面的向前移动,后面的也可以跟上,最后真正起作用的只是首和尾的距离。

#include <stdio.h>
#include <cstring>
#include <algorithm>
const int MAXN = 1005;
int T,n,a[MAXN];


template<typename _t>
inline _t read(){
    _t x=0,f=1;
    char ch=getchar();
    for(;ch>'9'||ch<'0';ch=getchar())if(ch=='-')f=-f;
    for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+(ch^48);
    return x*f;
}

int main(){
    T=read<int>();
    while(T--){
        n=read<int>();
        for(int i=1;i<=n;i++)a[i]=read<int>();
        std :: sort(&a[1],&a[n+1]);
        int xor_sum=0;
        for(int i=n;i>=1;i-=2){
            if(i==1)xor_sum^=a[i]-1;
            else xor_sum^=a[i]-a[i-1]-1;
        }
        if(xor_sum==0)printf("Bob will win\n");
        else printf("Georgia will win\n");
    }
}





基于python实现的粒子群的VRP(车辆配送路径规划)问题建模求解+源码+项目文档+算法解析,适合毕业设计、课程设计、项目开发。项目源码已经过严格测试,可以放心参考并在此基础上延申使用,详情见md文档 算法设计的关键在于如何向表现较好的个体学习,标准粒子群算法引入惯性因子w、自我认知因子c1、社会认知因子c2分别作为自身、当代最优解和历史最优解的权重,指导粒子速度和位置的更新,这在求解函数极值问题时比较容易实现,而在VRP问题上,速度位置的更新则难以直接采用加权的方式进行,一个常见的方法是采用基于遗传算法交叉算子的混合型粒子群算法进行求解,这里采用顺序交叉算子,对惯性因子w、自我认知因子c1、社会认知因子c2则以w/(w+c1+c2),c1/(w+c1+c2),c2/(w+c1+c2)的概率接受粒子本身、当前最优解、全局最优解交叉的父代之一(即按概率选择其中一个作为父代,不加权)。 算法设计的关键在于如何向表现较好的个体学习,标准粒子群算法引入惯性因子w、自我认知因子c1、社会认知因子c2分别作为自身、当代最优解和历史最优解的权重,指导粒子速度和位置的更新,这在求解函数极值问题时比较容易实现,而在VRP问题上,速度位置的更新则难以直接采用加权的方式进行,一个常见的方法是采用基于遗传算法交叉算子的混合型粒子群算法进行求解,这里采用顺序交叉算子,对惯性因子w、自我认知因子c1、社会认知因子c2则以w/(w+c1+c2),c1/(w+c1+c2),c2/(w+c1+c2)的概率接受粒子本身、当前最优解、全局最优解交叉的父代之一(即按概率选择其中一个作为父代,不加权)。
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