Playing With Stones
大意：n堆石子，每堆a1,a2....an个，两人轮流操作，每次选一堆，减少至少一个至多石子数一般下取整个。不能拿的人输。给定n,a,问先手必胜/必败
打表求得一堆的sg函数：
0 1 0 2 1 3 0 4 2 5 1 6 3 7 0 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15
不难发现其中有‘’1 2 3 5.....15"
即sg(2n) = n
考虑sg(2n + 1) 为0 1 0 2 1 3 0 4 2 5 1 6 3 7
恰好是原数列，即sg(2n+1) = sg(n) = sg([(2n+1)/2])
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <algorithm> 6 #include <queue> 7 #include <vector> 8 #include <cmath> 9 #define min(a, b) ((a) < (b) ? (a) : (b)) 10 #define max(a, b) ((a) > (b) ? (a) : (b)) 11 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a)) 12 inline void swap(long long &a, long long &b) 13 { 14 long long tmp = a;a = b;b = tmp; 15 } 16 inline void read(long long &x) 17 { 18 x = 0;char ch = getchar(), c = ch; 19 while(ch < '0' || ch > '9') c = ch, ch = getchar(); 20 while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); 21 if(c == '-') x = -x; 22 } 23 24 const long long INF = 0x3f3f3f3f; 25 const long long MAXN = 100 + 10; 26 27 long long t,n,a; 28 29 long long sg(long long x) 30 { 31 if(x == 0) return 0; 32 return (x & 1) == 0 ? x/2 : sg(x/2); 33 } 34 35 int main() 36 { 37 read(t); 38 for(;t;--t) 39 { 40 read(n);long long tmp = 0; 41 for(register long long i = 1;i <= n;++ i) 42 { 43 read(a); 44 tmp ^= sg(a); 45 } 46 if(tmp) printf("YES "); 47 else printf("NO "); 48 } 49 return 0; 50 }