威佐夫博弈

普通威佐夫博弈：

兩種操作：一、同時在兩堆上取相同的個數。二、在某一堆上取任意個數。（每次取不為0）

$a[n]=n\alpha ,b[n]=a[n]+n,\alpha =\frac{1+\sqrt{5}}{2}$，這個時候是可以確定必敗狀態的。

這里就不詳細展開證明了。

威佐夫博弈拓展：

兩種操作：一、同時在兩堆分別取$x,y，\mid x?y\mid <=k$。二、在某一堆上取任意個數。（每次取不為0）

對于給定的限定$k$，于是題目判定變成了$a[n]=n\alpha ,b[n]=a[n]+(k+1)n$，帶入$k$等于0的時候也就是普通情況下成立，

所以我們可以先解得$\alpha$，然后再通過二分去尋找是否有符合要求的$n$，這個時候$\alpha$的有關方程是$\frac{1}{\alpha }+\frac{1}{\alpha +k+1}=1$。

即可解的方程然后帶入驗證即可。

取(2堆)石子游戲（普通威佐夫）

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>#define mp make_pair #define pb push_back #define endl '\n' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using namespace std;typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii;const double pi = acos(-1.0); const double eps = 1e-7; const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int a, b;while(cin >> a >> b && (a + b)) {map<int, int> MP;if(a < b) swap(a, b);int k = (a - b) * ((1 + sqrt(5)) / 2);if(k == b) {puts("0");continue;}puts("1");for(int i = 1; i <= b; i++) {int n = a - i, m = b - i;int k = (n - m) * ((1 + sqrt(5)) / 2);if(k == m) {printf("%d %d\n", m, n);}}for(int i = 1; i <= a; i++) {int n = a - i, m = b;if(n < m) swap(n, m);int k = (n - m) * ((1 + sqrt(5)) / 2);if(MP[m] == n) continue;MP[m] = n;if(k == m) {printf("%d %d\n", min(n, m), max(n, m));}}for(int i = 1; i <= b; i++) {int n = a, m = b - i;if(n < m) swap(n, m);int k = (n - m) * ((1 + sqrt(5)) / 2);if(MP[m] == n) continue;MP[m] = n;if(k == m) {printf("%d %d\n", m, n);}}}return 0; }

Slime and Stones（拓展威佐夫）

/*Author : lifehappy */ #pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h>#define mp make_pair #define pb push_back #define endl '\n' #define mid (l + r >> 1) #define lson rt << 1, l, mid #define rson rt << 1 | 1, mid + 1, r #define ls rt << 1 #define rs rt << 1 | 1using namespace std;typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii;const double pi = acos(-1.0); const double eps = 1e-7; const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x; }ll calc(ll n, ll k) {double A = 1, B = k - 2, C = -k;double a = (sqrt(B * B - 4 * A * C) - B) / (2 * A);ll l = 1, r = 1e9;while(l < r) {ll Mid = l + r >> 1;ll now = Mid * a + k * Mid;if(now >= n) r = mid;else l = mid + 1;}ll now = l * a + k * l;return now == n ? l : -1; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = read();while(T--) {ll a = read(), b = read(), k = read() + 1;if(a < b) swap(a, b);ll n = calc(a, k);if(n == -1) puts("1");else {if(a - b == n * k) puts("0");else puts("1");}}return 0; }

總結

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