codeforces_622C. 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.

Examples

input

6 41 2 1 1 3 51 4 12 6 23 4 13 4 2

output

26-14

这道题两个做法吧，把重复的数字处理掉，这是一种，还有线段树的做法。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>using namespace std;int a[200010];int b[200010];//处理重复数字int solve(int l,int r,int x){    int bx,i;    for(i=l;i<=r;)    {        if(b[i]<=r)bx=b[i];        else bx=r;        if(a[bx]!=x)return bx;        i=bx+1;        //printf("i:%d\n",i);    }    return -1;}int main(){    int i,j,n,m,l,r,x;    scanf("%d%d",&n,&m);    for(i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    b[n]=n;    for(i=n-1;i>=1;i--)    {        if(a[i]==a[i+1])b[i]=b[i+1];        else b[i]=i;    }    for(i=1;i<=m;i++)    {        scanf("%d%d%d",&l,&r,&x);        printf("%d\n",solve(l,r,x));    }    return 0;}

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>using namespace std;//线段树const int MAXNODE = 200010;struct NODE{    int l,r,ma,mi;}node[MAXNODE<<2];int a[MAXNODE];int pushup(int i){    if(node[i].l==node[i].r)return 1;    node[i].ma=max(node[i<<1].ma,node[(i<<1)+1].ma);    node[i].mi=min(node[i<<1].mi,node[(i<<1)+1].mi);}int build(int i,int l,int r){    node[i].l=l;    node[i].r=r;    int mid=l+r>>1;    if(l==r)    {        node[i].ma=node[i].mi=a[node[i].l];        return 1;    }    build(i<<1,l,mid);    build((i<<1)+1,mid+1,r);    pushup(i);    return 1;}int ans,f=0;int query(int l,int r,int i,int x){    if(f)return 1;    if(node[i].l==node[i].r)    {        if(node[i].ma!=x)        {            ans= node[i].l;            f=1;        }        else ans= -1;        return 1;    }    int mid=node[i].l+node[i].r>>1;    if(node[i].ma!=x||node[i].mi!=x)    {        if(r<=mid)query(l,r,i<<1,x);        else if(l>mid)query(l,r,(i<<1)+1,x);        else        {            query(l,mid,i<<1,x);            query(mid+1,r,(i<<1)+1,x);        }        return 1;    }    else    {        ans=-1;        return 1;    }}int main(){    int n,m;    scanf("%d%d",&n,&m);    int i;    for(i=1;i<=n;i++)scanf("%d",&a[i]);    build(1,1,n);    while(m--)    {        f=0;        int l,r,x;        scanf("%d%d%d",&l,&r,&x);        query(l,r,1,x);        printf("%d\n",ans);    }    return 0;}

下面是两种做法的时空对比，第一个是线段树。