codeforces 622C C. Not Equal on a Segment

C. Not Equal on a Segment

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2

output

题意：给你一个数组，然后m条询问，在l[i]到r[i]之间是否存在一个数不等于x[i]，存在输出这个数的位置，不然输出-1;

思路：开一个fa数组，记录与这个数一直不间断相同的数的位置，相当于把相同的数合并，在查找的时候就省事了，开始直接暴力是过不了的；

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N=1e6+5;
int n,m,a[N],fa[N],l,r,x;
int solve(int le,int ri,int ans)
{
int fx;
for(int i=le;i<=ri;)
{
if(fa[i]<=ri)fx=fa[i];
else fx=ri;
if(a[fx]!=ans)return fx;
i=fx+1;
}
return -1;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
fa[n]=n;
for(int i=n-1;i>0;i--)
{
if(a[i]==a[i+1])fa[i]=fa[i+1];
else fa[i]=i;
}
while(m--)
{
scanf("%d%d%d",&l,&r,&x);
printf("%d\n",solve(l,r,x));
}
return 0;
}

巴特西

codeforces 622C C. Not Equal on a Segment

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