CF#52 C Circular RMQ （线段树区间更新）

Description

You are given circular array a₀, a₁, ..., a_n - 1. There are two types of operations with it:

• inc(lf, rg, v) — this operation increases each element on the segment [lf, rg] (inclusively) by v;
• rmq(lf, rg) — this operation returns minimal value on the segment [lf, rg] (inclusively).

Assume segments to be circular, so if n = 5 and lf = 3, rg = 1, it means the index sequence: 3, 4, 0, 1.

Write program to process given sequence of operations.

Input

The first line contains integer n (1 ≤ n ≤ 200000). The next line contains initial state of the array: a₀, a₁, ..., a_n - 1 ( - 10⁶ ≤ a_i ≤ 10⁶), a_i are integer. The third line contains integer m (0 ≤ m ≤ 200000), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf, rg (0 ≤ lf, rg ≤ n - 1) it means rmq operation, it contains three integers lf, rg, v (0 ≤ lf, rg ≤ n - 1; - 10⁶ ≤ v ≤ 10⁶) — inc operation.

Output

For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Sample Input

Input

4
1 2 3 4
4
3 0
3 0 -1
0 1
2 1

Output

1
0
0

题意非常好理解。

假设a>b的话，就查0~b和a~n-1，其余的就是线段树区间更新模板，推断m个询问里是否存在c，详见代码，非常好理解。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <iostream>
#define lson o << 1, l, m
#define rson o << 1|1, m+1, r
using namespace std;
typedef __int64 LL;
const __int64 MAX =  9223372036854775807;
const int maxn = 200010;
int n, a, q, c, b;
char str[1200];
LL mi[maxn<<2], add[maxn<<2];
void up(int o) {
    mi[o] = min(mi[o<<1], mi[o<<1|1]);
}
void down(int o) {
    if(add[o]) {
        add[o<<1] += add[o];
        add[o<<1|1] += add[o];
        mi[o<<1] += add[o];
        mi[o<<1|1] += add[o];
        add[o] = 0;
    }
}
void build(int o, int l, int r) {
    if(l == r) {
        scanf("%I64d", &mi[o]);
        return;
    }
    int m = (l+r) >> 1;
    build(lson);
    build(rson);
    up(o);
}
void update(int o, int l, int r) {
    if(a <= l && r <= b) {
        add[o] += c;
        mi[o] += c;
        return ;
    }
    down(o);
    int m = (l+r) >> 1;
    if(a <= m) update(lson);
    if(m < b ) update(rson);
    up(o);
}
LL query(int o, int l, int r) {
    if(a <= l && r <= b) return mi[o];
    down(o);
    int m = (l+r) >> 1;
    LL res = MAX;
    if(a <= m) res = query(lson);
    if(m < b ) res = min(res, query(rson));
    return res;
}
int main()
{
    scanf("%d", &n);
    build(1, 0, n-1);
    scanf("%d", &q); getchar(); while(q--) {
        gets(str);
        if(sscanf(str,"%d %d %d", &a, &b, &c) == 3)  {
            if(a <= b) update(1, 0, n-1);
            else {
                int tmp = b;
                b = n-1; update(1, 0, n-1);
                a = 0, b = tmp; update(1, 0, n-1);
            }
        } else {
            LL ans ;
            if(a <= b) ans = query(1, 0, n-1);
            else {
                int tmp = b;
                b = n-1; ans = query(1, 0, n-1);
                a = 0, b = tmp; ans = min(ans, query(1, 0, n-1));
            }
            printf("%I64d
", ans);
        }
    }
    return 0;
}