Finished, a few days before the DEF edu will not say, now playing face it qwq
In fact, just to finish the sentence would be a 1min D, 3min will be the E
We found that the size of the \ (\ ge 3 \) imbalance collection, which has a size of at least \ (2 \) subset is unbalanced.
Proof, found that for a size of \ (2 \) is set if and only if the balance of the number of two digits cross null (for an arbitrary, at least in this one a number is not \ (0 \) ).
A wave of evidence to the contrary, if the size is not large collection \ (2 \) imbalance collection, then any two digit number of the cross is empty, then the large collection is balanced, contradictory.
Therefore, only you need to consider the size of \ (2 \) collection.
This can be done a simple tree line, each segment tree maintenance \ (ans \) indicates that the interval of the answer, \ (mn_i \) indicates that the interval of the \ (i \) bits have the minimum number of values. See pushup specific code.
Time complexity \ (O ((n-Q + \ log n-) \ log a_i) \) .
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int maxn=200020,pw[9]={1,10,100,1000,10000,100000,1000000,10000000,100000000};
#define MP make_pair
#define PB push_back
#define lson o<<1,l,mid
#define rson o<<1|1,mid+1,r
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define ROF(i,a,b) for(int i=(a);i>=(b);i--)
#define MEM(x,v) memset(x,v,sizeof(x))
inline ll read(){
char ch=getchar();ll x=0,f=0;
while(ch<'0' || ch>'9') f|=ch=='-',ch=getchar();
while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
return f?-x:x;
}
int n,m,a[maxn];
struct node{
ll ans,mn[9];
node operator+(const node &nd)const{
node s;
s.ans=min(ans,nd.ans);
FOR(i,0,8) s.ans=min(s.ans,mn[i]+nd.mn[i]),s.mn[i]=min(mn[i],nd.mn[i]);
return s;
}
}seg[maxn*4];
inline int get(int x,int id){return x/pw[id]%10;}
void build(int o,int l,int r){
if(l==r){
seg[o].ans=1e18;
FOR(i,0,8) seg[o].mn[i]=get(a[l],i)?a[l]:1e18;
return;
}
int mid=(l+r)>>1;
build(lson);build(rson);
seg[o]=seg[o<<1]+seg[o<<1|1];
}
void update(int o,int l,int r,int p,int v){
if(l==r){
seg[o].ans=1e18;
FOR(i,0,8) seg[o].mn[i]=get(v,i)?v:1e18;
return;
}
int mid=(l+r)>>1;
if(mid>=p) update(lson,p,v);
else update(rson,p,v);
seg[o]=seg[o<<1]+seg[o<<1|1];
}
node query(int o,int l,int r,int ql,int qr){
if(l>=ql && r<=qr) return seg[o];
int mid=(l+r)>>1;
if(mid<ql) return query(rson,ql,qr);
if(mid>=qr) return query(lson,ql,qr);
return query(lson,ql,qr)+query(rson,ql,qr);
}
int main(){
n=read();m=read();
FOR(i,1,n) a[i]=read();
build(1,1,n);
while(m--){
int op=read(),x=read(),y=read();
if(op==1) update(1,1,n,x,y);
else{
ll ans=query(1,1,n,x,y).ans;
printf("%lld\n",ans>=2e9?-1ll:ans);
}
}
}