time limit per test4 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.
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.
【题解】
给你n个数字;
m个询问li,ri;
要让你在[li,ri]这个区间里面找到下标对i,j;
使得a[i]xor a[i+1] xor a[i+2]..xor a[j] == k;
让你输出在li,ri内这样的i,j对的个数;
n=10W;
m=100W;
每个数字ai最大为100W为非负数；
设sum[i]表示前i个数字的异或值;
离线处理询问；左端升序排;左端相同右端升序排;
然后从第一个询问开始处理;
设区间为L..R;
则i从L->R递增flag[sum[i]]
然后遇到一个sum[i]则递增答案flag[k^sum[i]];
假设k^sum[i] = sum[x] (x小于i）；
则有sum[x]^sum[i] = k;
而sum[i]^sum[x]实际上就是a[x+1]^a[x+2]..^a[i];
所以这个方式是可行的；
这样我们就能找出所有的点对了；
然后因为询问经过排序处理;
所以相邻询问的l和r和我们刚处理过的L,R是很接近的；
如果l
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson L,m,rt<<1
#define rson m+1,R,rt<<1|1
#define LL long long
using namespace std;
const int MAXN = 209999;
const int MAX_SIZE = 1009999;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
struct abc
{
int l,r,id;
};
int n,m,k,sum[MAXN];
LL ans[MAXN];
int flag[MAX_SIZE*2] = {0};
abc Q[MAXN];
void input_LL(LL &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)) t = getchar();
LL sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
void input_int(int &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)) t = getchar();
int sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
bool cmp(abc a,abc b)
{
if (a.l/400!=b.l/400)
return a.l/400<b.l/400;
else
return a.r < b.r;
}
int main()
{
//freopen("F:\rush.txt", "r", stdin);
input_int(n);input_int(m);input_int(k);
for (int i = 1;i <= n;i++)
{
int x;
input_int(x);
sum[i] = sum[i-1] ^ x;
}
for (int i = 1;i <= m;i++)
input_int(Q[i].l),input_int(Q[i].r),Q[i].l--,Q[i].id = i;
sort(Q+1,Q+1+m,cmp);
int L,R;
LL s = 0;
L = Q[1].l,R=Q[1].r;
for (int i = L;i <= R;i++)
{
s+= flag[k^sum[i]];
flag[sum[i]]++;
}
ans[Q[1].id] = s;
for (int i = 2;i <= m;i++)
{
int l = Q[i].l,r=Q[i].r;
while (L>l)
{
L--;
s+=flag[k^sum[L]];
flag[sum[L]]++;
}
while (L<l)
{
flag[sum[L]]--;
s-=flag[k^sum[L]];
L++;
}
while (R>r)
{
flag[sum[R]]--;
s-=flag[k^sum[R]];
R--;
}
while (R<r)
{
R++;
s+=flag[k^sum[R]];
flag[sum[R]]++;
}
ans[Q[i].id] = s;
}
for (int i = 1;i <= m;i++)
printf("%I64d
",ans[i]);
return 0;
}