是是非非

nim博弈的基础上增加了修改，a ^ a = 0。
所以每次修改的时候我们先异或上a[x]再异或上y,最后把a[x]变成y。
判断异或和是不是0即可。

#include <bits/stdc++.h>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

#ifdef LOCAL
#define debug(x) cout << "[" __FUNCTION__ ": " #x " = " << (x) << "]\n"
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
#define hash_ 1000000009
#define Continue(x) { x; continue; }
#define Break(x) { x; break; }
const int mod = 998244353;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
#define gc p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;
inline int read(){ static char buf[1000000], *p1 = buf, *p2 = buf; register int x = false; register char ch = gc; register bool sgn = false; while (ch != '-' && (ch < '0' || ch > '9')) ch = gc; if (ch == '-') sgn = true, ch = gc; while (ch >= '0'&& ch <= '9') x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc; return sgn ? -x : x; }
ll fpow(ll a, int b, int mod) { ll res = 1; for (; b > 0; b >>= 1) { if (b & 1) res = res * a % mod; a = a * a % mod; } return res; }
int a[N];
int main()
{
#ifdef LOCAL
    freopen("E:/input.txt", "r", stdin);
#endif
    int n, q;
    cin >> n >> q;
    int sum = 0;
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]), sum ^= a[i];
    while (q--)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        sum = sum ^ a[x] ^ y;
        a[x] = y;
        if (sum != 0)
            puts("Kan");
        else
            puts("Li");
    }
    return TIME;
}