CF 617E【莫队求区间异或和】

E. XOR and Favorite Number
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
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.

Input

The first line of the input contains integers nm 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.

 

题目大意:
给定长度为n的序列,m个询问和常数K
多个询问L,R,求有多少对(i,j)满足L<=i<=j<=R且i到j异或和为K

分析:
这题妙啊
总结经验
1.看到多个询问L,R的要考虑到莫队
2.看到区间异或和的要想到转化成前缀和,变成Si^Sj==K的简化问题
3.看到异或的时候要考虑到,经过异或权值范围会比数据中给得大(异或一下权值就不是原范围了)
3.看到权值不是很大的情况下,考虑能不能用桶
4.看到求多少对的问题要注意开long long
5.区间异或和转换成前缀和后,实际取值中i为L-1<=i<R,要特判加上L-1的贡献(就类似于树上莫队特判lca)

                                      转载自acha

2017-04-06

Select Code
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=4e5+5;
typedef long long ll;
struct block{int l,r,id;}Q[N];
int n,m,K,bsize,a[N];
ll nowans,ans[N],cnt[N*10];
inline int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
bool operator <(const block &a,const block &b){
	return a.l/bsize!=b.l/bsize?a.l/bsize<b.l/bsize:a.r<b.r;
}
inline void ins(int x){
	nowans+=cnt[K^x];
	cnt[x]++;
}
inline void del(int x){
	cnt[x]--;
	nowans-=cnt[K^x];
}
int main(){
	n=read();m=read();K=read();bsize=sqrt(n+0.5);
	for(int i=1;i<=n;i++) a[i]=read(),a[i]^=a[i-1];
	for(int i=1;i<=m;i++) Q[i].l=read(),Q[i].r=read(),Q[i].id=i;
	sort(Q+1,Q+m+1);
	int l=1,r=0;
	for(int i=1;i<=m;i++){
		while(l>Q[i].l) ins(a[--l]);
		while(l<Q[i].l) del(a[l++]);
		while(r<Q[i].r) ins(a[++r]);
		while(r>Q[i].r) del(a[r--]);
		ans[Q[i].id]=nowans+cnt[K^a[l-1]];
	}
	for(int i=1;i<=m;i++) printf("%lld\n",ans[i]);
	return 0;
}
 

 

#include<cstdio>
#include<cmath>
#include<algorithm>
#define EF if(ch==EOF) return EOF;
#define EX if(ch==EOF) return x*f;
using namespace std;
const int N=4e5+5;
struct node{int l,r,id;}q[N];
int n,m,k,bsize,a[N];
long long nowans,ans[N],cnt[N*10];
inline int read(){
    int x=0,f=1;char ch=getchar();EF
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();EF}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();EX}
    return x*f;
}
inline bool cmp(const node &a,const node &b){
    return a.l/bsize!=b.l/bsize?a.l/bsize<b.l/bsize:a.r<b.r;
}
inline void ins(int x){
    nowans+=cnt[k^x];
    cnt[x]++;
}
inline void del(int x){
    cnt[x]--;
    nowans-=cnt[k^x];
}
int main(){
    n=read();m=read();k=read();
    for(int i=1;i<=n;i++) a[i]=read(),a[i]^=a[i-1];
    for(int i=1;i<=m;i++) q[i].l=read(),q[i].r=read(),q[i].id=i;
    bsize=sqrt(n);
    sort(q+1,q+m+1,cmp);
    int l=q[1].l,r=q[1].r;
    for(int i=l;i<=r;i++) ins(a[i]);
    ans[q[1].id]=nowans+cnt[k^a[l-1]];
    for(int i=2;i<=m;i++){
        while(q[i].l<l) ins(a[--l]);
        while(q[i].r>r) ins(a[++r]);
        while(q[i].l>l) del(a[l++]);
        while(q[i].r<r) del(a[r--]);
        ans[q[i].id]=nowans+cnt[k^a[l-1]];
    }
    for(int i=1;i<=m;i++) printf("%I64d\n",ans[i]);
    return 0;
}

 

referance:

http://codeforces.com/blog/entry/22971(官方解题报告)

http://blog.csdn.net/qq978874169/article/details/51241737(莫队算法入门)

posted @ 2017-03-07 09:17  神犇(shenben)  阅读(629)  评论(0编辑  收藏  举报