T1.
Meaning of the questions:
A given n number of sequence a , a of each operation may be a XOR value becomes the number of the entire sequence. How many times will need to find at least a become a target sequence b, can not be completed output - 1
answer:
XOR problem (bit computing problems) will not be a pure analog (in fact, know also write positive solution ), it is finished to do more of this title and thought feel familiar with the operation.
A nature: a number of different or the same or twice the original number after number, enum prove it on the line.
The nature of this problem we can see that in fact the sequence is a head switching sequence n + 1, to achieve the desired sequence, or -1 output. The unequal exchange a [i] and b [i] is connected to the side, i.e., the number of switching blocks Unicom edges. Jumping to a different number of blocks to be added Unicom 1, n + 1, then a single count different from a communication block. Finally, the answer is not difficult to find: the communication block number + the number of edges -1, because the n + 1 is not operating, but we count it in the calculation of the time.
Code:
#include<bits/stdc++.h> #define For(i,l,r) for(int i=l;i<=r;i++) #define ll long long using namespace std; const int M=1e5+5; int n,a[M],b[M],x[M],y[M],vis[M],cnt,ans; map<int,int>d; vector<int>Q[M]; inline int read(){ int f=1,sum=0; char ch=getchar(); while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} while(isdigit(ch)){sum=(sum<<1)+(sum<<3)+(ch^48);ch=getchar();} return f*sum; } inline void dfs(int x){ vis[x]=1; For(i,0,Q[x].size()-1){ if(!vis[Q[x][i]]){ dfs(Q[x][i]); } } } int main(){ n=read(); For(i,1,n){ a[i]=read();a[n+1]^=a[i]; if(!d[a[i]]) d[a[i]]=++cnt; } if(!d[a[n+1]]) d[a[n+1]]=++cnt; For(i,1,n){ b[i]=read();b[n+1]^=b[i]; if(!d[b[i]]) d[b[i]]=++cnt; } if(!d[b[n+1]]) d[b[n+1]]=++cnt; memcpy(x,a,sizeof(x));memcpy(y,b,sizeof(y)); sort(x+1,x+n+2);sort(y+1,y+n+2); For(i,1,n) if(x[i]!=y[i]){printf("-1");return 0;} For(i,1,n+1) a[i]=d[a[i]],b[i]=d[b[i]]; For(i,1,n) if(a[i]!=b[i]) ans++,Q[a[i]].push_back(b[i]); if(a[n+1]!=b[n+1]) Q[a[n+1]].push_back(b[n+1]); For(i,1,cnt) if(Q[i].size()&&!vis[i]) dfs(i),ans++; printf("%d",ans-1); return 0; }