[BZOJ4888] [TJOI2017] xor (tree-like array)

Title Description

Xiao Ming in Garidon Middle school recently fell in love with the maths contest, and many maths competitions are sequential and related to the sequence. So for a sequence, find out all of their continuous and, xiaoming feel very simple. But today Xiao Ming encountered a sequence and the problem, this topic not only requires you to quickly find out all the continuous and, but also to quickly find out these successive and different values. Xiaoming soon to find out all the continuous and, but Xiao Ming to test you, in not telling the continuous and the case, let you quickly be realistic sequence all consecutive and the XOR value.

Input/output format

Input format:

The first line enters an n, which indicates that the sequence of numbers in the second row of input n number a1,a2...an represents this sequence

0<=A1,A2,... an,0<=a1+a2...+an<=10^6

Output format:

Outputs all successive and XOR values of this sequence.

Input and Output sample input sample #: Copy

31 2 3

Output Example # #: Replication

0

Description

"Sample Interpretation"

Sequence 1 2 3 has 6 consecutive and they are 1 2 3 3 5 6, then 1 XOR 2 XOR 3 XOR 3 XOR 5 xor 6 = 0

"Data Range"

For 20% of data, 1<=n<=100

For 100% of data, 1<=n <= 10^5

Interval and conversion to prefix and the difference, the tree-like array of the discussion can be.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 #defineRep (i,l,r) for (int i=l; i<=r; i++)5 using namespacestd;6 7 Const intn=100100, m=1000000;8 intflag,cnt,c[m+5][2],s[n],a[n],ans;9 intpw[ +],n;Ten  One voidAddintXintY) { while(x<=m) c[x][y]++,x+= (x& (-x)); } A intQueryintXinty) { -    intsum=0; while(x) sum+=c[x][y],x-= (x& (-x)); -    returnsum; the } -  - intMain () { -scanf"%d",&n); +Rep (I,1, N) scanf ("%d", &s[i]), s[i]+=s[i-1]; -pw[0]=1; Rep (I,1, -) pw[i]=pw[i-1]*2; +Rep (I,0, -) A       if(pw[i]<=S[n]) { atMemset (c,0,sizeof(c)); -flag=0; Add1,0); -Rep (J,1, N) { -             inttmp=s[j]&Pw[i]; -             if(TMP) Cnt=query (a[j]+1,0) +query (1000001,1)-query (a[j]+1,1); -             ElseCnt=query (a[j]+1,1) +query (1000001,0)-query (a[j]+1,0); in             if(cnt%2==1) flag^=1; -Add (a[j]+1,(BOOL) tmp); to             if(TMP) a[j]|=Pw[i]; +          } -          if(flag) ans|=(Pw[i]); the       } *printf"%d\n", ans); $    return 0;Panax Notoginseng}

[BZOJ4888] [TJOI2017] xor (tree-like array)