【HDOJ6610】Game（序列带修莫队）

题意：有n堆石子，第n堆有a[i]个，A先选择一个范围【L,R】，B选择一个子区间【l,r】，之后照nim游戏的规则进行

现在有询问与操作

每次询问B在给定的【L,R】内有多少种子区间的取法使得A必胜

每次操作会交换第x堆和第x+1堆石头

0<=a[i]<=1e6，n,m<=1e5

思路：

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 //typedef pair<ll,ll>P;
 #define N  3000010
 #define M  100010
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=0x7fffffff;
       int inf=1e9;
       int dx[]={-,,,};
       int dy[]={,,-,};

 struct Q
 {
     int l,r,cur,id;
 }q[M];

 struct P
 {
     int x,pre,now;
 }p[M];

 int a[M],b[M],v[M],cnt[N],pos[M];
 ll sum,ans[M];

 bool cmp(Q a,Q b)
 {
    return pos[a.l]<pos[b.l]
         ||pos[a.l]==pos[b.l]&&pos[a.r]<pos[b.r]
         ||pos[a.l]==pos[b.l]&&pos[a.r]==pos[b.r]&&a.cur<b.cur;
 }

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void go(int x)
 {
     if(v[x])
     {
          cnt[b[x]]--;
          sum-=cnt[b[x]];
     }
      else
      {
          sum+=cnt[b[x]];
          cnt[b[x]]++;
      }
     v[x]^=;
 }

 void update(int x)
 {
     if(!x) return;
     if(v[x])
     {
         go(x);
         b[x]^=a[x];
         swap(a[x],a[x+]);
         b[x]^=a[x];
         go(x);
     }
      else
      {
          b[x]^=a[x];
         swap(a[x],a[x+]);
         b[x]^=a[x];
      }
 }

 int main()
 {
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         mem(cnt,);
         mem(v,);
         mem(a,);
         mem(b,);
         int S=max(,(int)pow(n,2.0/));
         a[]=b[]=; pos[]=;
         rep(i,,n+)
         {
             a[i]=read();
             b[i]=b[i-]^a[i];
             pos[i]=(i-)/S;
         }
         int l1=,l2=;
         rep(i,,m)
         {
             int op=read();
             if(op==)
             {
                 int x=read(),y=read();
                 q[++l1].l=x;
                 q[l1].r=y+;
                 q[l1].id=l1;
                 q[l1].cur=l2;
             }
             else
             {
                 int x=read();
                 p[++l2].x=x+;
             }

         }
         sort(q+,q+l1+,cmp);
         int L=,R=,T=;
         sum=;
         rep(i,,l1)
         {
             while(T<q[i].cur)
             {
                 T++;
                 update(p[T].x);
             }

             while(T>q[i].cur)
             {
                 update(p[T].x);
                 T--;
             }

             while(L>q[i].l) go(--L);
             while(L<q[i].l) go(L++);
             while(R>q[i].r) go(R--);
             while(R<q[i].r) go(++R);
             ll len=R-L+;
             ans[q[i].id]=len*(len-)/-sum;
         }
         rep(i,,l1) printf("%I64d\n",ans[i]);
     }

     return ;
 }