【LOJ】#2210. 「HNOI2014」江南乐

LOJ#2210. 「HNOI2014」江南乐

感觉是要推sg函数

发现(lfloor frac{N}{i} floor)只有(O(sqrt{N}))种取值

考虑把这些取值都拿出来，能取到这个值的(i)是一个区间([l,r])

如果(r - l + 1 = 1)，那么直接算这个数的答案即可（(lfloor frac{N}{i} floor)的石子有奇数堆还是偶数堆，(lfloor frac{N}{i} floor + 1)的石子有奇数堆还是偶数堆，异或起来即可）

如果(r - l + 1 > 1)，证明这个区间里既有奇数又有偶数

其中(lfloor frac{N}{i} floor + 1)有(N - ilfloor frac{N}{i} floor)堆

(lfloor frac{N}{i} floor)有(i - (N - ilfloor frac{N}{i} floor))堆

由于(N)和(lfloor frac{N}{i} floor)奇偶性确定了，我们只要枚举两种不同奇偶性的(i)计算两种情况的游戏值就可以了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define ba 47
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int T,F;
int sg[100005],N;
unordered_map<int,int> zz;
void Process() {
    for(int i = 0 ; i < F ; ++i) {
	sg[i] = 0;
    }
    for(int x = F ; x <= 100000 ; ++x) {
	zz.clear();
	for(int i = 2 ; i <= x ; ++i) {
	    int r = x / (x / i);
	    int t = x / i;
	    if(r - i + 1 == 1) {
		int k = 0;
		if((x % r) & 1) k ^= sg[x / i + 1];
		if((r - (x % r)) & 1) k ^= sg[x / i];
		zz[k] = 1;
	    }
	    else {
		if(t % 2 == 0) {
		    if(x & 1) {
			zz[sg[x / i + 1] ^ sg[x / i]] = 1;
			zz[sg[x / i + 1]] = 1;
		    }
		    else zz[sg[x / i]] = 1;
		}
		else {
		    if(x & 1) {
			zz[sg[x / i + 1] ^ sg[x / i]] = 1;
			zz[sg[x / i]] = 1;
		    }
		    else zz[sg[x / i + 1]];
		}
	    }
	    i = r;
	}
	while(zz.count(sg[x])) sg[x]++;
    }
}
void Solve() {
    int a,ans = 0;
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
	read(a);
	ans ^= sg[a];
    }
    if(ans) putchar('1'),space;
    else putchar('0'),space;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(T);read(F);
    Process();
    for(int i = 1 ; i <= T ; ++i) {
	Solve();
    }
    return 0;
}