Bob has a favorite number?k?and?a_i?of length?n. Now he asks you to answer?m?queries. Each query is given by a pair?l_i?and?r_i?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?a_i,?a_i?+?1,?...,?a_jis 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?a_i?(0?≤?a_i?≤?1?000?000)?— Bob's array.

Then?m?lines follow. The?i-th line contains integers?l_i?and?r_i?(1?≤?l_i?≤?r_i?≤?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

題意：
詢問區(qū)間內(nèi)異或和剛好為k的字段個數(shù)。
思路：
莫隊+前綴和。
這個前綴和比較套路，用的是前綴異或和。
字段【l,r】的異或和就是pre[r]^pre[l-1],
這種情況下我們在莫隊的過程中記錄l，r的pre[i]出現(xiàn)的次數(shù)，就可以完成更新了。

注意當L<q[i].l時，要先讓記錄per[L]出現(xiàn)次數(shù)的個數(shù)減一。 #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<cstdio> #include<cstring> #include<cmath> #include<ctime> #define fuck(x) cout<<#x<<" = "<<x<<endl; #define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl; #define ls (t<<1) #define rs ((t<<1)+1) using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 100086; const int maxm = 1000086; const int inf = 2.1e9; const ll Inf = 999999999999999999; const int mod = 1000000007; const double eps = 1e-6; const double pi = acos(-1); int num[maxm*2],pre[maxn],a[maxn];struct node{int l,r;int id; }q[maxn]; ll ans[maxn]; ll anss; int block;bool cmp(node a,node b){if(a.l/block!=b.l/block){return a.l<b.l;}return a.r<b.r; }int main() {int n,m,k;scanf("%d%d%d",&n,&m,&k);for(int i=1;i<=n;i++){scanf("%d",&a[i]);pre[i]=pre[i-1]^a[i];}block=sqrt(n);for(int i=1;i<=m;i++){scanf("%d%d",&q[i].l,&q[i].r);q[i].id=i;}sort(q+1,q+1+m,cmp);int L=1,R=0;anss=0;num[0]=1;for(int i=1;i<=m;i++){while(L<q[i].l){int t=k^pre[L-1];num[pre[L-1]]--;///注意語句順序anss-=num[t];L++;}while(R>q[i].r){int t=k^pre[R];num[pre[R]]--;anss-=num[t];R--;}while(L>q[i].l){L--;int t=k^pre[L-1];anss+=num[t];num[pre[L-1]]++;}while(R<q[i].r){R++;int t=k^pre[R];anss+=num[t];num[pre[R]]++;}ans[q[i].id]=anss;}for(int i=1;i<=m;i++){printf("%lld\n",ans[i]);}return 0; } View Code

轉(zhuǎn)載于:https://www.cnblogs.com/ZGQblogs/p/10863851.html

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