HDU5875 - Function
做法:st表+二分的经典题。不能使用数学函数log,否则会tle,需要预处理
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define MP make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fr first
#define sc second
typedef long long ll;
const int N= 2e5 + 7;
using namespace std;
int n,m;
ll a[N], st[N][50],Log[N];void init() {Log[1] = 0;for(int i = 2; i <= n; ++i) Log[i] = Log[i>>1] + 1;for(int i = 0; i <= n; ++i) st[i][0] = a[i];for(int j = 1; (1<<j) <= n; ++j){for(int i = 1; i+(1<<j)-1 <= n; ++i) {st[i][j] = min(st[i][j-1], st[i+(1<<(j-1))][j-1]);}}
}
inline ll ask(int u,int v) {int k = Log[v-u+1];return min(st[u][k], st[v-(1<<k)+1][k]);
}int fd(int l,int r,ll x) {int s = r + 1;while(l <= r) {int mid = (l+r)>>1;if(ask(l,mid) <= x) r = mid - 1, s = mid;else l = mid + 1;}return s;
}
ll solve(int l,int r) {ll now = st[l][0];if(l == r) return now;l++;while(l<=r && now){l = fd(l,r,now);if(l <= r) {now %= st[l][0]; ++l;}}return now;
}
int main() {int T;scanf("%d",&T);while(T--) {scanf("%d",&n);rep(i,1,n) scanf("%lld",&a[i]);init();scanf("%d",&m);rep(i,1,m) {int l,r;scanf("%d%d",&l,&r);printf("%lld\n",solve(l,r));}}return 0;
}