$ \color{#0066ff}{ 題目描述 }$

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?

\(\color{#0066ff}{輸入格式}\)

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.

\(\color{#0066ff}{輸出格式}\)

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

\(\color{#0066ff}{輸入樣例}\)

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

\(\color{#0066ff}{輸出樣例}\)

Bob will win Georgia will win

\(\color{#0066ff}{數(shù)據(jù)范圍與提示}\)

none

\(\color{#0066ff}{題解}\)

移動一個棋子，與左邊的距離減小，與右邊的距離增大

可以抽象成幾堆石子，從一堆拿到另一堆，這就是階梯NIM了

#include <cctype> #include <cstdio> #include <algorithm>#define LL long long LL in() {char ch; LL x = 0, f = 1;while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));return x * f; } const int maxn = 2050; int a[maxn], b[maxn]; int main() {for(int T = in(); T --> 0;) {int n = in();for(int i = 1; i <= n; i++) a[i] = in();std::sort(a + 1, a + n + 1);int ans = 0;for(int i = n; i >= 1; i -= 2)ans = ans ^ (a[i] - a[i - 1] - 1) ; puts(!ans? "Bob will win" : "Georgia will win");}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/olinr/p/10491697.html

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