Hackerrank XOR Subsequences
2016-12-15 18:19
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求连续子串的Xor和出现次数最多的Xor和
Xoryi=xAi=(Xorx−1i=1Ai)Xor(Xoryi=1Ai)
最后发现其实就是个卷积形式
上FWT即可
中间处理过程可能会爆int 所以加上取模会免掉不必要的麻烦
Xoryi=xAi=(Xorx−1i=1Ai)Xor(Xoryi=1Ai)
最后发现其实就是个卷积形式
上FWT即可
中间处理过程可能会爆int 所以加上取模会免掉不必要的麻烦
#include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<cmath> using namespace std; char c; inline void read(int&a) {a=0;do c=getchar();while(c<'0'||c>'9');while(c<='9'&&c>='0')a=(a<<3)+(a<<1)+c-'0',c=getchar();} int A[100001],B[100001]; const int Mod=(int)1e9+7,Rev=(Mod+1)/2; int W[1<<16]; inline int add(int a,int b) { a+=b; if(a>=Mod)a-=Mod; if(a<0)a+=Mod; return a; } inline int mul(int a,int b) { return a*1ll*b%Mod; } int main() { int n; read(n); A[0]++; for(int i=1;i<=n;i++)read(W[i]),W[i]^=W[i-1],A[W[i]]++; for(int i=1;i<(1<<16);i*=2) for(int j=0;j<(1<<16);j+=2*i) for(int k=0;k<i;k++) { int x=A[j+k],y=A[i+j+k]; A[j+k]=add(x,y); A[i+j+k]=add(x,-y); } for(int i=0;i<(1<<16);i++) A[i]=mul(A[i],A[i]); for(int i=1;i<(1<<16);i*=2) for(int j=0;j<(1<<16);j+=2*i) for(int k=0;k<i;k++) { int x=A[j+k],y=A[i+j+k]; A[j+k]=mul(Rev,add(x,y)); A[j+k+i]=mul(Rev,add(x,-y)); } A[0]-=n+1; int L=0; for(int i=1;i<=n;i++) if(A[i]>A[L])L=i; printf("%d %d\n",L,A[L]>>1); return 0; }
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