[bzoj2124]等差子序列(hash+树状数组)

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #define LL long long  //别开unsigned
 5 const LL mod=(LL)1e9+7;  //非洲人老老实实去找别人家的素数
 6 int n;
 7 LL pw[10010];
 8 LL bit1[10010],bit2[10010];
 9 int a[10010];
10 /*int binpr(int x){
11     int sta[10]={0},top=0;
12     while(x){
13         top++;
14         if(x&1)sta[top]=1;
15     }
16     ++top;
17     while(--top){
18         printf("%d",sta[top]);
19     }
20     return 0;
21 }*/
22 int min(int a,int b){
23     return a<b?a:b;
24 }
25 int lb(int x){
26     return x&(-x);
27 }
28 LL q1(int x){
29     LL ans=0;
30     int i=x;
31     while(i){
32         ans=(ans+(bit1[i]*pw[x-i])%mod)%mod;
33         i-=lb(i);
34     }
35     return ans;
36 }
37 LL q2(int x){
38     LL ans=0;
39     int i=x;
40     while(i){
41         ans=(ans+(bit2[i]*pw[x-i])%mod)%mod;
42         i-=lb(i);
43     }
44     return ans;
45 }
46 LL c1(int x){
47     int i=x;
48     while(i<=n){
49         bit1[i]=(bit1[i]+pw[i-x])%mod;
50         i+=lb(i);
51     }
52     return 0;
53 }
54 LL c2(int x){
55     int i=x;
56     while(i<=n){
57         bit2[i]=(bit2[i]+pw[i-x])%mod;
58         i+=lb(i);
59     }
60     return 0;
61 }
62 LL qr1(int l,int r){
63     LL p=q1(l-1),q=q1(r);
64     return ((q-p*pw[r-l+1])%mod+mod)%mod;
65 }
66 LL qr2(int l,int r){
67     LL p=q2(l-1),q=q2(r);
68     return ((q-p*pw[r-l+1])%mod+mod)%mod;
69 }
70 int main(){
71     int t;
72     scanf("%d",&t);
73     pw[0]=1;
74     for(int i=1;i<=10001;i++)pw[i]=(pw[i-1]*(LL)2)%mod;
75     while(t--){
76         scanf("%d",&n);
77         for(int i=1;i<=n;i++)scanf("%d",&a[i]);//坑点：按照这种算法以为是在线，BUT！多组数据！千万别在线做得出解就立马跳出！最起码先读进来
78         memset(bit1,0,sizeof(bit1));
79         memset(bit2,0,sizeof(bit2));
80         int note=0;
81         for(int i=1;i<=n;i++){
82             int x=a[i];
83             int len=min(n-x,x-1);
84             if(len && qr1(x-len,x-1)!=qr2(n-x-len+1,n-x)){
85                 note=1;
86                 puts("Y");
87                 break;
88             }
89             c1(x);
90             c2(n-x+1);
91         }
92         if(!note)puts("N");
93     }
94     return 0;
95 }