Codeforces 272C Dima and Staircase【线段树】

C. Dima and Staircase

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an(1 ≤ a1 ≤ a2 ≤ ... ≤ an).

Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:

• the bottom of the box touches the top of a stair;
• the bottom of the box touches the top of a box, thrown earlier.

We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.

You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 ≤ ai ≤ 109; ai ≤ ai + 1).

The next line contains integer m (1 ≤ m ≤ 105) — the number of boxes. Each of the following m lines contains a pair of integers wi, hi(1 ≤ wi ≤ n; 1 ≤ hi ≤ 109) — the size of the i-th thrown box.

The numbers in the lines are separated by spaces.

Output

Print m integers — for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Examples

input

51 2 3 6 641 13 11 14 3

output

1346

input

31 2 321 13 1

output

13

input

1151 21 101 101 101 10

output

13132333

Note

The first sample are shown on the picture.

题目大意：

我们初始的时候有长度为N的一系列高度为a【i】的方块。

然后有Q个方块需要下落，下落的左端在位子1，每一次下落一个长度为wi，高度为hi的方块，问每一次这个方块下落到达的位子高度。

具体参考一下样例就懂了。

思路：

线段树区间更新，区间查询一下最大值即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;#define ll __int64ll a[150000];ll n;/***********************/#define lson l,m,rt*2#define rson m+1,r,rt*2+1ll tree[100001*4];ll flag[100001*4];void pushup(ll rt){    tree[rt]=max(tree[rt<<1],tree[rt<<1|1]);}void pushdown(ll l,ll r,ll rt){    if(flag[rt])    {        ll m=(l+r)/2;        flag[rt*2]=flag[rt*2+1]=flag[rt];        tree[rt*2]=flag[rt];        tree[rt*2+1]=flag[rt];        flag[rt]=0;    }}void build( ll l ,ll r , ll rt ){    flag[rt]=0;    if( l == r )    {        tree[rt]=0;        flag[rt]=0;    }    else    {        ll m = (l+r)>>1 ;        build(lson) ;        build(rson) ;        pushup(rt) ;    }}void update(ll L,ll R,ll c,ll l,ll r,ll rt){    if(L<=l&&r<=R)    {        tree[rt]=c;        flag[rt]=c;        return ;    }    pushdown(l,r,rt);    ll m=(l+r)/2;    if(L<=m)    {        update(L,R,c,lson);    }    if(m<R)    {        update(L,R,c,rson);    }    pushup(rt);}ll query(ll L,ll R,ll l,ll r,ll rt){    if(L<=l&&r<=R)    {        return tree[rt];    }    pushdown(l,r,rt);    ll ans=0,m=(l+r)/2;    if(L<=m)ans=max(ans,query(L,R,lson));    if(R>m)ans=max(ans,query(L,R,rson));    pushup(rt);    return ans ;}/***********************/int main(){    while(~scanf("%I64d",&n))    {        for(ll i=1; i<=n; i++)scanf("%I64d",&a[i]),a[i]--;        build(1,n,1);        for(ll i=1; i<=n; i++)update(i,i,a[i],1,n,1);        ll q;scanf("%I64d",&q);        while(q--)        {            ll Len,h;            scanf("%I64d%I64d",&Len,&h);            ll maxn=query(1,Len,1,n,1)+1;            printf("%I64d\n",maxn);            update(1,Len,maxn+h-1,1,n,1);        }    }}