Sereja and Suffixes
64-bit integer IO format: %I64d Java class name: (Any)
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each li.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the array elements.
Next m lines contain integers l1, l2, ..., lm. The i-th line contains integer li (1 ≤ li ≤ n).
Output
Print m lines — on the i-th line print the answer to the number li.
Sample Input
10 10
1 2 3 4 1 2 3 4 100000 99999
1
2
3
4
5
6
7
8
9
10
6
6
6
6
6
5
4
3
2
1
Source
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 100010; 4 struct Query { 5 int idx,ask; 6 } Q[maxn]; 7 bool visib[maxn]; 8 int n,m,d[maxn],ret[maxn]; 9 bool cmp(const Query &a,const Query &b) { 10 return a.ask > b.ask; 11 } 12 int main() { 13 scanf("%d %d",&n,&m); 14 for(int i = 0; i < n; ++i) 15 scanf("%d",d+i); 16 for(int i = 0; i < m; ++i) { 17 scanf("%d",&Q[i].ask); 18 Q[i].idx = i; 19 } 20 sort(Q,Q+m,cmp); 21 for(int i = 0,j = n-1,cnt = 0; i < m && j >= 0; ++i) { 22 while(j >= 0 && Q[i].ask - 1 != j) { 23 if(!visib[d[j]]) { 24 cnt++; 25 visib[d[j]] = true; 26 } 27 j--; 28 } 29 if(j >= 0 && Q[i].ask - 1 == j) { 30 if(!visib[d[j]]) { 31 cnt++; 32 visib[d[j]] = true; 33 } 34 ret[Q[i].idx] = cnt; 35 } 36 } 37 for(int i = 0; i < m; ++i) 38 printf("%d\n",ret[i]); 39 return 0; 40 }