CodeFroce Round 340 div2 E XOR and Favorite Number【莫队算法】
2017-08-09 12:28
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题面:
Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri and asks you to count the number of pairs of integers i and j, such that l ≤ i ≤ j ≤ r and the xor of the numbers ai, ai + 1, …, aj is equal to k.Input
The first line of the input contains integers n, m and k (1 ≤ n, m ≤ 100 000, 0 ≤ k ≤ 1 000 000) — the length of the array, the number of queries and Bob’s favorite number respectively.
The second line contains n integers ai (0 ≤ ai ≤ 1 000 000) — Bob’s array.
Then m lines follow. The i-th line contains integers li and ri (1 ≤ li ≤ ri ≤ n) — the parameters of the i-th query.
Output
Print m lines, answer the queries in the order they appear in the input.
Examples
input
6 2 3
1 2 1 1 0 3
1 6
3 5
output
7
0
input
5 3 1
1 1 1 1 1
1 5
2 4
1 3
output
9
4
4
Note
In the first sample the suitable pairs of i and j for the first query are: (1, 2), (1, 4), (1, 5), (2, 3), (3, 6), (5, 6), (6, 6). Not a single of these pairs is suitable for the second query.
In the second sample xor equals 1 for all subarrays of an odd length.
大致思路:
对于数组中的每一个数求前缀和,公式为:pref[i]=pref[i-1]^a[i]
然后根据异或的性质可以知道:
a[i] ^ a[i+1]^…..^a[j]=pref[j]-pref[i-1]
还有一个公式:
a^b=k 可以推出: a^k=b
这样就可以利用莫队算法了来解决这道题了。
首先需要开一个大小为2^20的桶(虽然题上说数据只到1e6,但异或之后的值是有可能比1e6大的,那就干脆开大一点)
然后bow[i]的意思是异或值为i的数在区间内出现了几次。这样修改区间的时候就可以利用这个桶对答案进行修改。具体见代码。
然后是关于莫队算法说一些自己的理解
莫队算法的核心是分块+提前知道询问区间。也就是这些询问操作都是离线的,中间不能对区间有修改。然后就是根据块将询问排序,然后就暴力开始让L和R变化到询问区间,这样使得减少L,R的无效操作从而加快查询效率。
代码:
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1<<20; struct node{ int l,r,id; }query[maxn];//储存询问的结构体数组 ll a[maxn],ans[maxn],Ans=0; int bow[maxn],size,pos[maxn],k; bool cmp(node a,node b)//根据分块进行排序 { if(pos[a.l]==pos[b.l]) return a.r<b.r; return pos[a.l]<pos[b.l]; } void add(int x) { Ans+=bow[a[x]^k];//利用公式可以得到在x为结束点,异或值为k的对数 bow[a[x]]++; } void del(int x) { bow[a[x]]--; Ans-=bow[a[x]^k]; } int main() { ios::sync_with_stdio(false); int n,m; cin>>n>>m>>k; size=sqrt(n); for(int i=1;i<=n;++i){ cin>>a[i]; a[i]=a[i]^a[i-1]; pos[i]=i/size;//分块 } for(int i=1;i<=m;++i){ cin>>query[i].l>>query[i].r; query[i].id=i; } bow[0]=1; sort(query+1,query+m+1,cmp); int L=1,R=0; for(int i=1;i<=m;++i){ while(L<query[i].l)//调整区间 { del(L-1);//这里的顺序是根据第二个公式得到 ++L; } while(L>query[i].l) { --L; add(L-1); } while(R<query[i].r) { ++R; add(R); } while(R>query[i].r) { del(R); --R; } ans[query[i].id]=Ans; } for(int i=1;i<=m;++i) cout<<ans[i]<<endl; return 0; }
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