hdu—6059

Kanade's trio

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 255    Accepted Submission(s): 74

Problem Description

Give you an array A[1..n]，you need to calculate how many tuples (i,j,k) satisfy that (i<j<k) and ((A[i] xor A[j])<(A[j] xor A[k]))

There are T test cases.

1≤T≤20

1≤∑n≤5∗105

0≤A[i]<230

 

Input

There is only one integer T on first line.

For each test case , the first line consists of one integer n ,and the second line consists of n integers which means the array A[1..n]

 

Output

For each test case , output an integer , which means the answer.

 

Sample Input

151 2 3 4 5

 

Sample Output

6

 

【分析】

字典树.....枚举a[j]，考虑1~j-1维护一棵字典树，j+1~n维护一棵字典树，每次计算a[j]的每一位对应的两棵树的每一层满足后缀树^(a[j]^(1<<k))>前缀树^(a[j]^(1<<k))的方案数，也就是当前位置如果是0，就计算后缀树当前层中1的数量和前缀树当前层中0的数量

【代码】

#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>#define X 33#define N 600000#define M 20000000using namespace std;int n, a[N], cnt, f[M][3], s[X+3], trans[M];long long ans_case, ans[X+3][3], g[M][3];void update(int pos,int x,int modi,int t,int dep)//当前节点pos,当前位01状态x,add/delete状态modi //前缀树和后缀树区分t(0/1)，层数dep {g[pos][t]+=modi;//正常更新当前节点数量 ans[dep][x]+=modi*g[trans[pos]][t^1];//计算dep层中，后缀树中当前节点为x，前缀树中对应x^1的方案数 }void add(int x,int t){int j=0, ct=0;while (x){s[++ct]=x&1;x>>=1;}for (int i=ct+1;i<=X;i++) s[i]=0;for (int i=X;i;i--){if (!f[j][s[i]]){f[j][s[i]]=++cnt;if (f[j][s[i]^1]){trans[cnt]=f[j][s[i]^1];//记录当前cnt节点在另一棵树中对应的节点编号，用于后面计算答案 trans[f[j][s[i]^1]]=cnt;}}j=f[j][s[i]];update(j,s[i]^t,1,t,i);}}void del(int x,int t){int j=0, ct=0;while (x){s[++ct]=x&1;x>>=1;}for (int i=ct+1;i<=X;i++) s[i]=0;for (int i=X;i;i--){j=f[j][s[i]];update(j,s[i]^t,-1,t,i);}}void solve(int x){int j=0, ct=0;while (x){s[++ct]=x&1;x>>=1;}for (int i=ct+1;i<=X;i++) s[i]=0;for (int i=X;i;i--) ans_case+=ans[i][s[i]];}int main(){int T_T;scanf("%d",&T_T);while (T_T--){cnt=ans_case=0;scanf("%d",&n);for (int i=1;i<=n;i++) scanf("%d",&a[i]);if (n==1 || n==2){cout<<0<<endl;continue;}//高中生的题目....总有些坑留在这 for (int i=1;i<=n;i++) add(a[i],1);del(a[1],1);add(a[1],0);for (int i=2;i<=n;i++){del(a[i],1);solve(a[i]);add(a[i],0);}cout<<ans_case<<endl;for (int i=1;i<=X;i++) ans[i][0]=ans[i][1]=0;for (int i=0;i<=cnt;i++) f[i][0]=f[i][1]=g[i][0]=g[i][1]=trans[i]=0;}return 0;}

标程

#include<bits/stdc++.h>#define fi first#define se second#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)#define per(i,j,k) for(int i=(int)j;i>=(int)k;i--)using namespace std;typedef long long LL;const int N=510000;inline void read(int &x){    x=0;char p=getchar();    while(!(p<='9'&&p>='0'))p=getchar();    while(p<='9'&&p>='0')x*=10,x+=p-48,p=getchar();}int a[N];int go[N*32][2],tot;LL ss[N*32];int num[N*32];int sum[N][32][2];int n,T;int main(){    read(T);    while(T--){        rep(i,1,tot)rep(j,0,1)go[i][j]=0;        rep(i,1,tot)ss[i]=num[i]=0;        rep(i,1,n)rep(j,0,29)rep(k,0,1)sum[i][j][k]=0;        tot=1;        read(n);        rep(i,1,n)read(a[i]);        rep(i,1,n){            rep(j,0,29)rep(k,0,1)sum[i][j][k]=sum[i-1][j][k];            rep(j,0,29){                int v=((a[i]&(1<<j))>0);                sum[i][j][v]++;            }        }        LL ans=0;        per(i,n,1){            int now=1;            per(j,29,0){                int v=((a[i]&(1<<j))>0);                if(go[now][v^1]){                    ans+=ss[go[now][v^1]]-num[go[now][v^1]]*1ll*sum[i][j][v];                }                if(!go[now][v])break;                now=go[now][v];            }            now=1;            per(j,29,0){                int v=((a[i]&(1<<j))>0);                if(!go[now][v])go[now][v]=++tot;                now=go[now][v];                ss[now]+=sum[i-1][j][v^1];                num[now]++;            }        }        cout<<ans<<endl;    }    //cerr<<clock()<<endl;    return 0;}