codeforces 52C Circular RMQ（线段树区间更新）【模板】

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 mlines 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).

Example

Input

41 2 3 443 03 0 -10 12 1

Output

100

 【题解】 裸的线段树区间更新，自zz了一下，把数组写错了，WA了好几次才过，现在附上AC代码。

 

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;const int N=200006;#define INF 0x3fffffff#define lson ans<<1,ll,mid#define rson ans<<1|1,mid+1,rr__int64 m,n,sum[N],mins[N<<2],lazy[N<<2];char str[200];__int64 min(__int64 x,__int64 y){    return x>y?y:x;}void pushup(int ans){    mins[ans] =min(mins[ans<<1],mins[ans<<1|1]);}void pushdown(int ans){    if(lazy[ans])    {        lazy[ans<<1]+=lazy[ans] ;        lazy[ans<<1|1]+=lazy[ans] ;        mins[ans<<1]+= lazy[ans];        mins[ans<<1|1]+=lazy[ans];        lazy[ans]=0;    }}void build_tree(int ans,int ll,int rr){    lazy[ans] =0;    if(ll == rr )    {        mins[ans] = sum[ll];        return ;    }    int mid = (ll + rr )>>1;    build_tree(lson);    build_tree(rson);    pushup(ans);}void update(int ans,int ll,int rr,int L,int R,int value){    if(L<=ll && R>= rr)    {        lazy[ans]+=value;        mins[ans]+=value;        return ;    }    pushdown(ans);    int mid = (ll+rr)>>1;    if(mid<R)//更新从mid+1到R的的区间        update(rson,L,R,value);    if(mid>=L)        update(lson,L,R,value);    pushup(ans);}__int64 query(int ans,int ll,int rr,int L,int R){    if(ll>=L && rr<= R)//是否在区间内    {        return mins[ans];    }    pushdown(ans);    int mid = (ll+rr)>>1;    __int64 ss=INF;    if(mid >= L)//往左走    {        ss=min(ss,query(lson,L,R));    }    if(mid<R)//往右走       ss = min(ss,query(rson,L,R));    return ss;}int main(){    while(~scanf("%I64d",&n))    {        int a,b,c;        n--;        for(int i=0;i<=n;i++)            scanf("%I64d",&sum[i]);        build_tree(1,0,n);        scanf("%I64d",&m);        getchar();        for(int i=1;i<=m;i++)        {            gets(str);            if(sscanf(str,"%d%d%d",&a,&b,&c)==3)            {                if(a>b)                {                    update(1,0,n,a,n,c);                    update(1,0,n,0,b,c);                }                else                    update(1,0,n,a,b,c);            }            else            {                if(a>b)                {                    __int64 ans=min(query(1,0,n,a,n),query(1,0,n,0,b));                    printf("%I64d\n",ans);                }                else                {                    printf("%I64d\n",query(1,0,n,a,b));                }            }        }    }    return 0;}