bzoj 2844 子集异或和名次
感谢：
http://blog.sina.cn/dpool/blog/s/blog_76f6777d0101d0mr.html
的讲解（特别是2^(n-m)的说明）。
1 /************************************************************** 2 Problem: 2844 3 User: idy002 4 Language: C++ 5 Result: Accepted 6 Time:308 ms 7 Memory:2052 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <iostream> 12 #define Mod 10086 13 #define N 100010 14 using namespace std; 15 16 int n, m, q; 17 int aa[N], bit[N]; 18 19 void gauss() { 20 int i=0; 21 for( int b=30; b>=0; b-- ) { 22 for( int j=i; j<n; j++ ) 23 if( (aa[j]>>b)&1 ) { 24 swap( aa[j], aa[i] ); 25 break; 26 } 27 if( (aa[i]>>b)&1 ) { 28 for( int j=i+1; j<n; j++ ) 29 if( (aa[j]>>b)&1 ) 30 aa[j] ^= aa[i]; 31 bit[i] = b; 32 i++; 33 } 34 } 35 m = i; 36 for( int i=0; i<m; i++ ) 37 for( int j=i-1; j>=0; j-- ) 38 if( (aa[j]>>bit[i])&1 ) 39 aa[j] ^= aa[i]; 40 } 41 int mpow( int a, int b ) { 42 int rt; 43 for( rt=1; b; b>>=1,a=(a*a)%Mod ) 44 if( b&1 ) rt=(rt*a)%Mod; 45 return rt; 46 } 47 int main() { 48 scanf( "%d", &n ); 49 for( int i=0; i<n; i++ ) 50 scanf( "%d", aa+i ); 51 scanf( "%d", &q ); 52 gauss(); 53 int cur = 0; 54 int rank = (q==0?0:1); 55 for( int i=0; i<m; i++ ) { 56 if( (cur^aa[i])<q ) { 57 cur ^= aa[i]; 58 rank = (rank + (1<<(m-1-i))%Mod) % Mod; 59 } 60 } 61 rank = (rank*mpow(2,n-m)+1) % Mod; 62 printf( "%d\n", rank ); 63 }
