2012-05-30 06:39:21Morris
[UVA][DP] 10404 - Bachet's Game
Problem B: Bachet's Game
Bachet's game is probably known to all but probably not by this name.
Initially there are n stones on the table. There are two
players Stan and Ollie, who move alternately. Stan always starts. The
legal moves consist in removing at least one but not more than
k stones from the table. The winner is the one to take the
last stone.
Here we consider a variation of this game. The number of stones that can be removed in a single move must be a member of a certain set of m numbers. Among the m numbers there is always 1 and thus the game never stalls.
Input
The input consists of a number of lines. Each line describes one game by a sequence of positive numbers. The first number is n <= 1000000 the number of stones on the table; the second number is m <= 10 giving the number of numbers that follow; the last m numbers on the line specify how many stones can be removed from the table in a single move.Input
For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly.Sample input
20 3 1 3 8 21 3 1 3 8 22 3 1 3 8 23 3 1 3 8 1000000 10 1 23 38 11 7 5 4 8 3 13 999996 10 1 23 38 11 7 5 4 8 3 13
Output for sample input
Stan wins Stan wins Ollie wins Stan wins Stan wins Ollie wins
狀態轉移的方程並不難理解, 贏會變輸, 輸會變成贏
#include <stdio.h>
char dp[1000005];
int main() {
int n, m, s[15], i, j;
while(scanf("%d %d", &n, &m) == 2) {
for(i = 0; i < m; i++)
scanf("%d", &s[i]);
dp[0] = 0;
for(i = 1; i <= n; i++) {
dp[i] = 0;
for(j = 0; j < m; j++) {
if(i-s[j] >= 0) {
if(dp[i-s[j]] == 0) {
dp[i] = 1;
break;
}
}
}
}
if(dp[n])
puts("Stan wins");
else
puts("Ollie wins");
}
return 0;
}