TOJ 2703: Cow Digit Game

2703: Cow Digit Game

Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte
Total Submit: 1            Accepted:1

Description

Bessie is playing a number game against Farmer John, and she wants you to help her achieve victory.

Game i starts with an integer N_i (1 <= N_i <= 1,000,000). Bessie goes first, and then the two players alternate turns. On each turn, a player can subtract either the largest digit or the smallest non-zero digit from the current number to obtain a new number. For example, from 3014 we may subtract either 1 or 4 to obtain either 3013 or 3010, respectively. The game continues until the number becomes 0, at which point the last player to have taken a turn is the winner.

Bessie and FJ play G (1 <= G <= 100) games. Determine, for each game, whether Bessie or FJ will win, assuming that both play perfectly (that is, on each turn, if the current player has a move that will guarantee his or her win, he or she will take it).

Consider a sample game where N_i = 13. Bessie goes first and takes 3, leaving 10. FJ is forced to take 1, leaving 9. Bessie takes the remainder and wins the game.

Input

* Line 1: A single integer: G

* Lines 2..G+1: Line i+1 contains the single integer: N_i

Output

* Lines 1..G: Line i contains "YES" if Bessie can win game i, and "NO" otherwise.

Sample Input

2
9
10

Sample Output

YES
NO

Hint

OUTPUT DETAILS:

For the first game, Bessie simply takes the number 9 and wins. For the second game, Bessie must take 1 (since she cannot take 0), and then FJ can win by taking 9.

Source

USACO Open 2009

简单的博弈，sg函数

把一个数减去这个数非0的最大值或者最小值，分析其必胜

#include <stdio.h>
#include <algorithm>
using namespace std;
const int N=1e6+;
int sg[N];
int main(){
for(int i=;i<=1e6;i++){
int ma=,mi=,t=i;
while(t){
    int b=t%;
    if(b){
    ma=max(b,ma);
    mi=min(b,mi);}
    t/=;
}
sg[i]=(sg[i-ma]&sg[i-mi])^;
}
int t;
scanf("%d",&t);
while(t--){
    int n;
    scanf("%d",&n);
    printf("%s",sg[n]?"YES\n":"NO\n");
}
return ;}