G - Circular RMQ

题意：
有两种操作，一种是环上的区间加，一种是环上的区间最值
题解：
线段树区间更新模板

#include<iostream>
#include <cstring>
#include<string>
#include <algorithm>
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 5000;
const ll inf = 1e18;
ll mi[maxn << 2], add[maxn << 2];
int n, q;
void PushUp(int rt)
{
    mi[rt] = min(mi[rt << 1], mi[rt << 1 | 1]);
}
void PushDown(int rt)
{
    if (add[rt])
    {
        mi[rt << 1] += add[rt];
        mi[rt << 1 | 1] += add[rt];
        add[rt << 1] += add[rt];
        add[rt << 1 | 1] += add[rt];
        add[rt] = 0;
    }
}
void Build(int l, int r, int rt = 1)
{
    add[rt] = 0;
    if (l == r)
    {
        cin >> mi[rt];
        return;
    }
    int mid = (l + r) >> 1;
    Build(lson);
    Build(rson);
    PushUp(rt);
}
void Update(int L, int R, int c, int l, int r, int rt)
{
    if (L <= l && r <= R)
    {
        mi[rt] += c;
        add[rt] += c;
        return;
    }
    int mid = (l + r) >> 1;
    PushDown(rt);
    if (L <= mid)
        Update(L, R, c, lson);
    if (mid < R)
        Update(L, R, c, rson);
    PushUp(rt);
}
ll Query(int L, int R, int l, int r, int rt)
{
    if (L <= l && r <= R)
        return mi[rt];
    int mid = (l + r) >> 1;
    ll ans = inf;
    PushDown(rt);
    if (L <= mid)
        ans = min(ans, Query(L, R, lson));
    if (mid < R)
        ans = min(ans, Query(L, R, rson));
    return ans;
}
int main()
{
    cin >> n;
    Build(1, n, 1);
    cin >> q;
    while (q--)
    {
        int l, r, c;
        cin >> l >> r;
        l++;
        r++;
        if (getchar() == ' ')
        {
            cin >> c;
            if (l <= r)
                Update(l, r, c, 1, n, 1);
            else
            {
                Update(1, r, c, 1, n, 1);
                Update(l, n, c, 1, n, 1);
            }
        }
        else
        {
            if (l <= r)
                cout << Query(l, r, 1, n, 1) << endl;
            else
                cout << min(Query(1, r, 1, n, 1), Query(l, n, 1, n, 1)) << endl;
        }
    }
}