275. To xor or not to xor
分析:
线性基模板题。
代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 5 const int N = 100 + 10; 6 const int m = 61; 7 LL a[N],b[N]; 8 int n; 9 10 void build() { 11 for (int i=1; i<=n; ++i) 12 for (int j=m; j>=0; --j) 13 if ((1LL<<j)&a[i]) { 14 if (b[j]) a[i] ^= b[j]; 15 else { 16 b[j] = a[i]; 17 for (int k=j-1; k>=0; --k) if (b[k]&&((1LL<<k)&b[j])) b[j] ^= b[k]; 18 for (int k=j+1; k<=m; ++k) if ((1LL<<j)&b[k]) b[k] ^= b[j]; 19 break; 20 } 21 } 22 23 } 24 void solve() { 25 LL ans = 0; 26 for (int i=0; i<=m; ++i) 27 if (b[i]) ans ^= b[i]; 28 cout << ans; 29 } 30 int main() { 31 scanf("%d",&n); 32 for (int i=1; i<=n; ++i) scanf("%I64d",&a[i]); 33 build(); 34 solve(); 35 return 0; 36 }