HDU 5875 Function

题意：给一个数组，然后m个查询，查询[L, R]中，f = a[L] % a[L + 1] ..... % a[R]，f的值

思路：因为a % b = a，当a < b时，所以只需使用线段树维护区间最小值，然后跳过被模数大于要模的数的区间

#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<string>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#include<iostream>
typedef long long ll;
const int maxn = 1e6 + 10;
const double eps = 1e-7;
const ll INF = 1e18;
using namespace std;

int tree[4 * maxn], a[maxn];
int ql, qr, ind;
int T, m, q, n;
int ans;

void update(int node, int l, int r) {
    if(l == r) { tree[node] = m; return ; }
    int mid = (l + r) >> 1;
    if(ind <= mid) update(node * 2, l, mid);
    else update(node * 2 + 1, mid + 1, r);
    tree[node] = min(tree[node * 2], tree[node * 2 + 1]);
}

void query(int node, int l, int r) {
    if(ql > qr) return ;
    if(tree[node] > ans) return ;
    if(l == r) { ans %= tree[node]; return ; }
    int mid = (l + r) >> 1;
    if(ql <= mid) query(node * 2, l, mid);
    if(qr > mid) query(node * 2 + 1, mid + 1, r);
}

int main()
{
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        for(ind = 1; ind <= n; ind++) {
            scanf("%d", &m);
            update(1, 1, n);
            a[ind] = m;
        }
        scanf("%d", &q);
        while(q--) {
            scanf("%d %d", &ql, &qr);
            ans = a[ql++];
            query(1, 1, n);
            printf("%d\n", ans);
        }
    }
    return 0;
}