FZU2072——二分——Count

Given an array of positive integers and m queries.Each query contains i, j, x, output the number of occurrences of x into the subarray Ai,Ai+1...,Aj.

Input

There are several cases. The first line of each case contains tow integers n, q(1<=n,q<=100000), indicating the array length and the number of queries.The second line contains n positive integers ai(1 <= ai <= 100000).Next q lines contain three positive integers i,j,x(1<=i<=j<=n).

Output

For each query output one line, the number of occurrences of x.

Sample Input

3 2 1 2 1 1 2 1 1 3 1

Sample Output

1 2

/*

lower_bound(  ,  , x)

是用二分从begin 到 end 找序号大于等于 x 的第一个值为y的迭代器下标

大意：在l到r之间询问k出现的次数

*/

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;



const int MAX = 1000100;

vector<int> G[MAX];

int a[MAX];



int  cal(int x, int y)

{   

    return lower_bound( G[y].begin(), G[y].end(), x) - G[y].begin();

}

int main()

{

int n, m;

int l, r, k;

while(~scanf("%d%d", &n, &m)){

    for(int i = 1; i <= n; i++)

        G[a[i]].clear();

    for(int i = 1; i <= n; i++){

        scanf("%d", &a[i]);

        G[a[i]].push_back(i);

    }

    for(int i = 1; i <= m; i++){

        scanf("%d%d%d", &l, &r, &k);

        printf("%d\n", cal(r + 1 , k) - cal(l , k));

    }

}

    return 0;

}