CodeForces 52C Circular RMQ (区间更新线段树)

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本文介绍了一种解决循环数组上的范围最小查询（RMQ）及区间增量操作的问题，通过使用线段树进行优化处理，实现了高效的查询和更新操作。

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C - Circular RMQ

Time Limit:3000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status

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

Output

1
0
0

#include<iostream>
#include<cstdio>
#define MAX 200005
#define INF 0x3fffffff
long long c,tree[4*MAX],add[4*MAX];
long long _min(long long a,long long b)
{
    return a<b?a:b;
}

void pushdown(int n)
{
    if(add[n]!=0)
    {
        add[2*n]+=add[n];
        add[2*n+1]+=add[n];
        tree[2*n]+=add[n];
        tree[2*n+1]+=add[n];
        add[n]=0;
    }
}

void pushup(int n)
{
    tree[n]=_min(tree[2*n],tree[2*n+1]);
}

void build(int s,int e,int n)
{
    add[n]=0;
    if(s==e)
    {
        scanf("%I64d",&tree[n]);
        return ;
    }
    int mid=(s+e)/2;
    build(s,mid,2*n);
    build(mid+1,e,2*n+1);
    pushup(n);
}

void modify(int a,int b,long long c,int s,int e,int n)
{
    if(a<=s&&e<=b)
    {
        add[n]+=c;
        tree[n]+=c;
        return ;
    }
    pushdown(n);
    int mid=(s+e)/2;
    if(a<=mid) modify(a,b,c,s,mid,2*n);
    if(mid<b) modify(a,b,c,mid+1,e,2*n+1);
    pushup(n);
}

long long query(int a,int b,int s,int e,int n)
{
    if(a<=s&&e<=b)
    {
        return tree[n];
    }
    pushdown(n);
    int mid=(e+s)/2;
    long long s2,s1;s1=s2=INF;
    if(a<=mid) s1=_min(s1,query(a,b,s,mid,2*n));
    if(mid<b) s2=_min(s2,query(a,b,mid+1,e,2*n+1));
    return _min(s1,s2);
}

int main()
{
    int n,m,i,a,b;
    char q;
    scanf("%d",&n);
    build(1,n,1);
    scanf("%d",&m);
    while(m--)
    {
        scanf("%d%d",&a,&b);
        q=getchar();
        if(q==' ')
        {
            scanf("%I64d",&c);
            if(a<=b)
            {
                modify(a+1,b+1,c,1,n,1);
            }
            else
            {
                  modify(a+1,n,c,1,n,1);
                  modify(1,b+1,c,1,n,1);
            }
        }
        else
       {
           if(a<=b) printf("%I64d\n",query(a+1,b+1,1,n,1) );
           else
           {
               long long s1=query(a+1,n,1,n,1);
               long long s2=query(1,b+1,1,n,1);
               printf("%I64d\n",_min(s1,s2) );
           }
       }
    }
    return 0;
}