POJ 1704 Georgia and Bob (博弈)
and poj 博弈
2023-09-11 14:17:18 时间
题意:给定一个格子,有一些位置有球,每次可以向前移动一些位置,但不能超过前一个小球,问谁会胜利。
析:由于两个人都是足够聪明,所以我们可以把每两个相邻的球分成一个组,因为先手移动前一个移动多少位置,后手也可以移动第后一个,
如果是奇数,可以在0号位置放一个,没有影响,然后就成一个Nim组合游戏了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2000 + 10; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int main(){ int T; cin >> T; while(T--){ scanf("%d", &n); vector<int> v; v.push_back(0); for(int i = 0; i < n; ++i){ scanf("%d", &m); v.push_back(m); } sort(v.begin(), v.end()); int ans = 0; for(int i = n; i > 0; i -= 2) ans ^= (v[i] - v[i-1] - 1); if(ans) puts("Georgia will win"); else puts("Bob will win"); } return 0; }
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