[POI2009]KAM-Pebbles BZOJ1115 [ 待填坑 ] 博弈

有N堆石子，除了第一堆外，每堆石子个数都不少于前一堆的石子个数。两人轮流操作每次操作可以从一堆石子中移走任意多石子，但是要保证操作后仍然满足初始时的条件谁没有石子可移时输掉游戏。问先手是否必胜。

感谢MT大牛翻译.

Sample OutputNIE TAKHint

Input

第一行u表示数据组数。对于每组数据，第一行N表示石子堆数，第二行N个数ai表示第i堆石子的个数(a1<=a2<=……<=an)。 1<=u<=10 1<=n<=1000 0<=ai<=10000

Output

u行，若先手必胜输出TAK，否则输出NIE。

Sample Input2 2 2 2 3 1 2 4

 

转换为 阶梯NIM游戏；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/


int T;
int n; int a[maxn];
int pp[maxn];

int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >> T;
	while (T--) {
		int ans = 0;
		cin >> n;
		for (int i = 1; i <= n; i++)cin >> a[i];
		for (int i = 1; i <= n; i++) {
			pp[i] = a[i] - a[i - 1];
		}
		for (int i = n; i >= 1; i -= 2) {
			ans ^= pp[i];
		}
		if (ans)cout << "TAK" << endl;
		else cout << "NIE" << endl;
	}
	return 0;
}