Problem:
Range Minimum Query (RMQ)

Write a program which manipulates a sequence A = {a0, a1, . . . , an-1} with the following operations:

find(s, t): report the minimum element in as, as+1, . . . ,at.

update(i, x): change ai to x.

Note that the initial values of ai (i = 0, 1, . . . , n−1) are 231-1.

Input

n q
com0 x0 y0
com1 x1 y1
...
comq−1 xq−1 yq−1

In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes update(xi, yi) and '1' denotes find(xi, yi).

Output

For each find operation, print the minimum element.

Constraints

1 ≤ n ≤ 100000

1 ≤ q ≤ 100000

If comi is 0, then 0 ≤ xi < n, 0 ≤ yi < 231-1.

If comi is 1, then 0 ≤ xi < n, 0 ≤ yi < n.

Sample Input 1

3 5
0 0 1
0 1 2
0 2 3
1 0 2
1 1 2

Sample Output 1

1
2

Sample Input 2

1 3
1 0 0
0 0 5
1 0 0

Sample Output 2

2147483647
5

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

void find(int a[],int x,int y){
    int i;
    int min=pow(2,31)-1;
    for(i=x;i<=y;i++){
        if(a[i]<min)min=a[i];
    }
    printf("%d\n",min);
}

int main(void){
    int i;
    
    int n;
    scanf("%d ",&n);
    
    int a[n];
    for(i=0;i<n;i++)a[i]=pow(2,31)-1;

    int q;
    scanf("%d\n",&q);
    
    int com,x,y;
    while(q--){
        scanf("%d %d %d\n",&com,&x,&y);
        switch(com){
            case 0:
                a[x]=y;
                break;
            case 1:
                find(a,x,y);
                break;
        }
    }

    return 0;
}