Not Equal on a Segment CodeForces - 622C (思维)

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.

Example
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

给出一个数组，让求出给定区间中和给定的值不同的数
想了快一个小时，没有什么想法。。。。。

题解：对于整个数组进行预处理，记录下它前面的第一个和它不等的值得下边，查找时从右往左找（写法不同，方向也不同），当前值如果和它不等则输出该值，如果和它相等则跳到它前面第一个和它不等的值。。

#include<stdio.h>
int a[250010];
int pre[250010];

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(i != 1)
                if(a[i]==a[i-1])
                    pre[i]=pre[i-1];
                else pre[i ] = i-1;
            else pre[i] = -1;   
        }

    int l,r,v;
    for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&l,&r,&v);
            int j=r;
            while(1)
                {
                    if(j<l)
                        {
                            printf("-1\n");
                            break;
                        }

                    if(a[j] != v)
                        {
                            printf("%d\n",j);
                            break;
                        }
                    else j=pre[j];
                }
        }
    return 0;
}