题目链接:http://codeforces.com/problemset/problem/617/E
4 seconds
256 megabytes
standard input
standard output
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.
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.
Print m lines, answer the queries in the order they appear in the input.
6 2 3
1 2 1 1 0 3
1 6
3 5
7
0
5 3 1
1 1 1 1 1
1 5
2 4
1 3
9
4
4
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.
题意:
给出一个序列,作m此查询,每次查询的内容为:在区间[l, r]内,有多少个子区间的异或和为k?
题解:
莫队算法:解决区间询问的离线方法,时间复杂度:O(n^1.5)。
代码如下:
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+;
const int maxn = 1e5+; int n, m, k, w, a[maxn];
LL sum, ans[maxn], c[];
//a[i]为前缀异或和,c[i]为在当前区间内,前缀异或和(从1开始)为i的个数。
//可知:a[l-1]^a[r] = val[l]^val[l+1]^………^val[r] struct node
{
int l, r, id;
bool operator<(const node &x)const{
if(l/w==x.l/w) return r<x.r;
return l/w<x.l/w;
}
}q[maxn]; void del(int i)
{
c[a[i]]--;
sum -= c[a[i]^k];
} void add(int i)
{
sum += c[a[i]^k];
c[a[i]]++;
} int main()
{
scanf("%d%d%d",&n,&m,&k);
for(int i = ; i<=n; i++)
{
scanf("%d",&a[i]);
a[i] ^= a[i-];
}
for(int i = ; i<=m; i++)
{
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id = i;
} w = sqrt(n);
sort(q+,q++m); int L = , R = ;
c[] = , sum = ;
for(int i = ; i<=m; i++)
{
while(L<q[i].l) del(L-), L++;
while(L>q[i].l) L--, add(L-);
while(R<q[i].r) R++, add(R);
while(R>q[i].r) del(R), R--;
ans[q[i].id] = sum;
} for(int i = ; i<=m; i++)
printf("%lld\n",ans[i]);
return ;
}