The game of Osho Gym

原题链接

• 题意：博弈，(n) 个游戏，每个游戏有 (b_i) 堆石子，每个人只能拿 (a_i) 的指数倍的数量，然后不能行动的人输。
• 题解：一开始莽规律，很容易的发现是和奇偶性改变有关，然后看出来如果 (a_i) 是奇数，那么 (b_i) 奇数则先手必胜为 (sg_i) 为 (1)，然后如果 (a_i) 是偶数，那么就乱找规律，其实应该看 (sg) 并且很明显是个 (nim) 游戏，所以应该异或每个游戏的 (sg) 值。
• 代码：

#include <iostream>
#include <bits/stdc++.h>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 2e5+9;
int a[N], b[N];
int SG[N];

int sg[2020][2202];
void solve() {
    int n;cin >> n;
    memset(SG, 0, sizeof SG);
    for (int i = 1; i <= n ;i ++) {
        cin >> a[i] >> b[i];
        int len =0;
        if (a[i] == 1||a[i] > b[i]) {
            len = b[i];
            SG[i] = b[i] & 1;
            continue;
        }
        ll d = a[i] + 1;
        if (b[i] % d == b[i]) {
            SG[i] = 2;
        } else {
            if (b[i] %d != 0) {
                b[i] %= d;
                SG[i] = b[i] & 1;
            }
        }
    }
    ll nim = 0;
    for (int i = 1; i <= n; i ++) {
        nim ^= SG[i];
    }
    if (nim)cout <<1 << endl;
    else cout << 2 << endl;
}
int tem[900];
void gogo() {
    for (int i = 6; i <= 500; i ++) {
        sg[i][0] = 0;
        cout << "i:" << i;
        for (int j = 1;j <= 500;j ++) {
            cout <<" j:" << j << endl;

            memset(tem, 0, sizeof tem);
            //cout << "j-1  " << j-1 << endl;
            tem[sg[i][j-1]]=1;
            for (int k = 1,d = i; d <= j; d *= d) {
                tem[sg[i][j-d]] = 1;
            }
            for (int k = 0; k <= j; k ++) {
                if (!tem[k]) {
                    sg[i][j] = k;break;
                    //cout << k << '
';break;
                }
            }
            
        }
        for (int j = 1; j <= 100; j ++) {
            cout << j<<"= " << sg[i][j] << "
";
        }cout << endl;
        getchar();
    }
}
int main() {
    //ios::sync_with_stdio(0);
    freopen("powers.in", "r", stdin);
//  freopen("output.out","w", stdout);

    //gogo();
    int t=1;cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}