HDU5875

HDU5875 - Function

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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
",solve(l,r));
        }
    }
    return 0;
}