【第400篇题解纪念2016年10月28日】【28.10%】【codeforces 617E】XOR and Favorite Number
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\n",ans[i]); return 0; }
- Codeforces 617E:XOR and Favorite Number 莫队算法
- 【莫队算法】[CodeForces - 617E/Round #340]XOR and Favorite Number
- 【CodeForces-617E】XOR and Favorite Number 莫队(好玩题)
- [CQOI 2018]异或序列&[Codeforces 617E]XOR and Favorite Number
- CodeForces - 617E:XOR and Favorite Number (莫队算法入门题)
- Codeforces 617 E. XOR and Favorite Number
- 【CF 617E】 XOR and Favorite Number (Mo's algorithm)
- CodeForces 617 E. XOR and Favorite Number
- 617E - XOR and Favorite Number
- CODEFORCES 340 XOR and Favorite Number 莫队模板题
- Codeforces Round #340 (Div. 2) E - XOR and Favorite Number (莫队算法)
- Codeforces Round #340 (Div. 2) E. XOR and Favorite Number(分块 (java))
- Codeforces Round #340 (Div. 2)E. XOR and Favorite Number 莫队算法
- codeforces XOR and Favorite Number (莫队分块)
- Codeforces Round #340 (Div. 2) E. XOR and Favorite Number —— 莫队算法
- [CF617E]XOR and Favorite Number
- Codeforces Round #340 (Div. 2) E. XOR and Favorite Number
- [CF617E]XOR and Favorite Number/[CQOI2018]异或序列
- XOR and Favorite Number
- Codeforces 617EXOR and Favorite Number