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
Copy
6 4 1 2 1 1 3 5 1 4 1 2 6 2 3 4 1 3 4 2
output
Copy
2 6 -1 4
题意:给你一个数组,给你m个询问,每个询问一个区间, 一个数值,在区间里哪一个位置不是这个数值,输出任意一个就行;
分析:
此题关键在于查询,如何优化查询,其实我们发现我让我们做的无非查询很简单,我们想一想头一个数要么是,要么不是,所以就查询一次就ok,我们可以用一个预处理fa[a[i]]表示与a[i]不同的第一个位置是多少,这样就可以优化查询了。
#include<stdio.h>
#include<string>
#include<string.h>
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const int MAXN=200000+5;//最大元素个数
const int mod=100000007;
int n;//元素个数
int fa[MAXN];
int a[MAXN];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=-1)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<n;i++)
{
if(a[i]!=a[i+1])
{
fa[i]=i+1;
}
else
{
int pre=i;
for(;a[i]==a[i+1];)
{
i++;
}
//cout<<pre<<" "<<i<<endl;
for(int j=pre;j<=i;j++)
fa[j]=i+1;
}
}
while(m--)
{
int x,y,v;
scanf("%d%d%d",&x,&y,&v);
int flag=0;
for(int i=x;i<=y;i++)
{
if(a[i]==v)
{
if(fa[i]>y||fa[i]==0) break;
printf("%d\n",fa[i]);
flag=1;
break;
}
else
{
printf("%d\n",i);
flag=1;
break;
}
}
if(flag==0)
printf("-1\n");
}
//printf("%lld\n",ans);
}
return 0;
}