C. Not Equal on a Segment
You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.
For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.
The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.
Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.
Output
Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value - 1 if there is no such number.
Examples
input
6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2
output
2
6
-1
4
1 #include<cstdio> 2 #include<cstring> 3 #include<map> 4 #include<stack> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 typedef long long ll; 9 const int maxn=2e5+5; 10 const int INF=0x3f3f3f3f; 11 int pos=0; 12 struct tree 13 { 14 int l,r,minval,maxval,minval_id,maxval_id; 15 }s[maxn<<2]; 16 int check1(int a,int b) 17 { 18 if(s[a].minval>s[b].minval) 19 a=b; 20 return a; 21 } 22 int check2(int a,int b) 23 { 24 if(s[a].maxval<s[b].maxval) 25 a=b; 26 return a; 27 } 28 void pushup(int rt) 29 { 30 if(s[rt].minval>s[rt<<1].minval)s[rt].minval=s[rt<<1].minval,s[rt].minval_id=s[rt<<1].minval_id; 31 if(s[rt].minval>s[rt<<1|1].minval)s[rt].minval=s[rt<<1|1].minval,s[rt].minval_id=s[rt<<1|1].minval_id; 32 if(s[rt].maxval<s[rt<<1].maxval)s[rt].maxval=s[rt<<1].maxval,s[rt].maxval_id=s[rt<<1].maxval_id; 33 if(s[rt].maxval<s[rt<<1|1].maxval)s[rt].maxval=s[rt<<1|1].maxval,s[rt].maxval_id=s[rt<<1|1].maxval_id; 34 } 35 void build(int l,int r,int rt) 36 { 37 s[rt].minval=INF,s[rt].maxval=-INF; 38 s[rt].l=l,s[rt].r=r; 39 if(l==r) 40 { 41 int t; 42 scanf("%d",&t); 43 s[rt].maxval=s[rt].minval=t; 44 pos++; 45 s[rt].minval_id=pos; 46 s[rt].maxval_id=pos; 47 return; 48 } 49 int m=(l+r)>>1; 50 build(l,m,rt<<1); 51 build(m+1,r,rt<<1|1); 52 pushup(rt); 53 } 54 int query1(int l,int r,int rt) 55 { 56 if(l==s[rt].l&&r==s[rt].r) 57 return rt; 58 int m=(s[rt].l+s[rt].r)>>1; 59 if(l>m) 60 return query1(l,r,rt<<1|1); 61 else if(r<=m) 62 return query1(l,r,rt<<1); 63 else 64 { 65 int t1=query1(l,m,rt<<1),t2=query1(m+1,r,rt<<1|1); 66 return check1(t1,t2); 67 } 68 } 69 int query2(int l,int r,int rt) 70 { 71 if(l==s[rt].l&&r==s[rt].r) 72 return rt; 73 int m=(s[rt].l+s[rt].r)>>1; 74 if(l>m) 75 return query2(l,r,rt<<1|1); 76 else if(r<=m) 77 return query2(l,r,rt<<1); 78 else 79 { 80 int t1=query2(l,m,rt<<1),t2=query2(m+1,r,rt<<1|1); 81 return check2(t1,t2); 82 } 83 } 84 int main() 85 { 86 int n,m; 87 scanf("%d%d",&n,&m); 88 build(1,n,1); 89 while(m--) 90 { 91 int l,r,x; 92 scanf("%d%d%d",&l,&r,&x); 93 int rt1=query1(l,r,1); 94 int rt2=query2(l,r,1); 95 bool key=0; 96 int ans; 97 if(s[rt1].minval!=x) 98 ans=s[rt1].minval_id,key=1; 99 if(s[rt2].maxval!=x) 100 ans=s[rt2].maxval_id,key=1; 101 key?printf("%d ",ans):puts("-1"); 102 } 103 return 0; 104 }