连续区间的最大公约数

连续区间的最大公约数

分析

 

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#define pa pair<int,int>
using namespace std;
typedef long long LL;
 
inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}
 
const int N = 200005;
int gcd(int a,int b) { return b == 0 ? a : gcd(b, a % b); }
struct Node {
    int len, g; LL cnt; vector< pa > pre, suf;
    Node() { cnt = 0; }
    Node (int v) {
        pre.push_back(pa(v, 1)); suf.push_back(pa(v, 1));
        g = v, cnt = 0, len = 1;
    }
}T[N << 2];
int a[N];
 
#define fi first
#define se second
#define mp make_pair
void Merge(vector<pa> &a,vector<pa> &b) {
    for (int i = 0; i < (int)b.size(); ++i) {
        pa p = b[i];
        if (p.fi % a.back().fi == 0) a.back().se += p.se;
        else a.push_back(pa(gcd(p.fi, a.back().fi), p.se));
    }
}
LL Calc(vector<pa> &a, vector<pa> &b,int g) {
    LL res = 0;
    int sz = b.size() - 1, sum = 0;
    for (int i = 0; i <= sz; ++i) sum += b[i].se;
    for (int i = 0; i < (int)a.size(); ++i) {
        pa p = a[i];
        while (sz >= 0 && gcd(p.fi, b[sz].fi) == g) sum -= b[sz --].se;
        res += 1ll * p.se * sum;
    }
    return res;
}
Node operator + (Node A, Node B) {
    Node res;
    res.g = gcd(A.g, B.g);
    res.cnt += (res.g == A.g ? A.cnt : 1ll * A.len * (A.len + 1) / 2);
    res.cnt += (res.g == B.g ? B.cnt : 1ll * B.len * (B.len + 1) / 2);
    res.len = A.len + B.len;
    res.pre = A.pre, res.suf = B.suf;
    Merge(res.pre, B.pre);
    Merge(res.suf, A.suf);
    res.cnt += Calc(A.suf, B.pre, res.g);
    return res;
}
void build(int l,int r,int rt) {
    if (l == r) { T[rt] = Node(a[l]); return ; }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1); build(mid + 1, r, rt << 1 | 1);
    T[rt] = T[rt << 1] + T[rt << 1 | 1];
}
Node query(int l,int r,int rt,int L,int R) {
    if (L <= l && r <= R) return T[rt];
    int mid = (l + r) >> 1;
    if (R <= mid) return query(l, mid, rt << 1, L, R);
    else if (L > mid) return query(mid + 1, r, rt << 1 | 1, L, R);
    else return query(l, mid, rt << 1, L, R) + query(mid + 1, r, rt << 1 | 1, L, R);
}
void solve() {
    int n = read(), m;
    for (int i = 1; i <= n; ++i) a[i] = read();
    build(1, n, 1);
    m = read(); Node ans;
    while (m --) {
        int l = read(), r = read();
        ans = query(1, n, 1, l, r);
        printf("%d %lld
", ans.g, 1ll * ans.len * (ans.len + 1) / 2 - ans.cnt);
    }
}
int main() {
    for (int T = read(), i = 1; i <= T; ++i) {
        printf("Case #%d:
", i);
        solve();
    }
    return 0;
}