linkkkk
题意：
Nim游戏，任取几堆石子使得先手必败，问有几种方案数。
思路：
跟上题类似
方案数就是2 自 由 元 个 数 2^{自由元个数}2
自由元个数
代码：
#include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<ll, ll>PLL; typedef pair<int, int>PII; typedef pair<double, double>PDD; typedef pair<string,string>PSS; #define I_int ll inline ll read(){ll x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;} inline void write(ll x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');} #define read read() #define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0) #define multiCase int T;cin>>T;for(int t=1;t<=T;t++) #define rep(i,a,b) for(int i=(a);i<=(b);i++) #define repp(i,a,b) for(int i=(a);i<(b);i++) #define per(i,a,b) for(int i=(a);i>=(b);i--) ll ksm(ll a, ll b,ll mod){ll res = 1;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;} const int maxn=2010,mod=1e9+7; const double pi = acos(-1); ll n,m=31; ll a[110][110],g[110]; int Gauss(int n,int m) { int r, c; for(r = 0, c = 0; c < n; c++) { int t = -1; for(int i = r; i < m; i++) { if(a[i][c]) { t = i; break; } } if(t==-1) continue; for(int i = c; i < n; i++)///交换 { swap(a[t][i], a[r][i]); } for(int i = r + 1; i < m; i++) { if(a[i][c]) for(int j = c; j <n; j++) { a[i][j] = a[i][j] ^ a[r][j]; } } r++; } return n-r; /*if(r < n) { for(int i = r; i < n; i++) { if(a[i][n]) return 2; } return 1; } for(int i = n - 1; i >= 0; i--) { for(int j = i + 1; j < n; j++) { if(a[i][j]) a[i][n] ^= a[j][n]; } } return 0;*/ } void solve() { cin >> n; for(int i=0;i<n;i++) cin>>g[i]; for(int i=0;i<m;i++) for(int j=0;j<n;j++) a[i][j]=(g[j]>>i)&1; int ans=Gauss(n,m); ll mod=1000007; cout<<ksm(2,ans,mod)<<endl; } int main() { int _;cin>>_; while(_--) solve(); return 0; }
