鏈接
題解
感覺這題也就的難度吧(雖然我也是看了題解纔會的)
如果我把數組排好序,前綴和記作,第個數爲
現在假設用前面個數能連續湊出到,那麼當的時候,就沒法湊出這個數了,連擊終止,答案就是;否則,我就能湊出,並且加入這個數之後,我能一直連續湊到
所以問題就轉化成查詢最小的使得
(我就想到這裏,然後就不會做了)
看了題解之後,注意到一個十分好的性質,那就是我可以不必通過連續枚舉來依次檢查是否。因爲會發現,如果滿足了,那麼就會增加,這個幾乎就是乘以了
這個時候,其實我不用非得去檢查,我完全可以直接跳到後面去,而只要不合法就會直接退出,合法的話就會大小乘以,這樣單次查詢的時間複雜度其實是三個(因爲使用了樹狀數組套線段樹)
代碼
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 200010
#define maxs 320
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<ll,ll> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
struct joinable_segment_tree
{
ll sum[maxn*150];
int ch[maxn*150][2], tot;
void clear(){tot=0;}
ll New(ll v=0)
{
tot++;
sum[tot]=v;
ch[tot][0]=ch[tot][1]=0;
return tot;
}
ll create(ll l, ll r, ll pos, ll v) //創建一條鏈
{
ll ret=New(v), o=ret;
while(l<r)
{
ll mid(l+r>>1);
if(pos<=mid)o=ch[o][0]=New(v), r=mid;
else o=ch[o][1]=New(v), l=mid+1;
}
return ret;
}
ll join(ll u, ll v)
{
if(!u or !v)return u|v;
ch[u][0]=join(ch[u][0],ch[v][0]);
ch[u][1]=join(ch[u][1],ch[v][1]);
sum[u]=sum[u]+sum[v];
return u;
}
ll qsum(ll o, ll l, ll r, ll _l, ll _r)
{
if(!o)return 0;
ll mid(_l+_r>>1), ans(0);
if(l<=_l and r>=_r)return sum[o];
if(l<=mid)ans+=qsum(ch[o][0],l,r,_l,mid);
if(r>mid)ans+=qsum(ch[o][1],l,r,mid+1,_r);
return ans;
}
}mori;
struct BIT
{
ll bit[maxn], n;
void init(ll N){n=N;for(ll i=1;i<=n;i++)bit[i]=0;}
ll lowbit(ll x){return x&(-x);}
void add(ll pos, ll v, ll num)
{
for(;pos<=n;pos+=lowbit(pos))
{
bit[pos] = mori.join( mori.create(1,2e5,num,v), bit[pos] );
}
}
ll sum(ll pos, ll l, ll r )
{
ll ans(0);
for(;pos;pos-=lowbit(pos))
{
ans += mori.qsum( bit[pos], l, r, 1, 2e5 );
}
return ans;
}
}bit;
int main()
{
de( (sizeof(mori)>>20) );
ll n=read(), Q=read(), i;
vector<ll> a(maxn);
rep(i,1,n)a[i]=read();
bit.init(2e5);
rep(i,1,n)bit.add(a[i],+a[i],i);
while(Q--)
{
ll type=read(), x=read(), y=read();
if(type==1)
{
bit.add(a[x],-a[x],x);
bit.add(y,+y,x);
a[x]=y;
}
else
{
ll now = 0;
while(now<2e5 and bit.sum(now+1,x,y)>now)
{
now = bit.sum(now+1,x,y);
}
if(now<2e5)printf("%lld\n",now+1);
else printf("%lld\n",bit.sum(2e5,x,y)+1);
}
}
return 0;
}