Educational Codeforces Round 72 (Rated for Div. 2)E. Sum Queries?（线段树区间合并）

https://codeforc.es/contest/1217/problem/E

建立9棵数位线段树维护区间最小值和次小值，建议用struct建树方便进行区间合并

 1 #define bug(x) cout<<#x<<" is "< 2 #define IO std::ios::sync_with_stdio(0)
 3 #include 
 4 #define iter ::iterator
 5 #define pa pair
 6 using namespace  std;
 7 #define ll long long
 8 #define mk make_pair
 9 #define pb push_back
10 #define se second
11 #define fi first
12 #define ls o<<1
13 #define rs o<<1|1
14 ll mod=998244353;
15 const int N=2e5+5,h=2e9+5;
16 int n,q;
17 struct node{
18     int mi[10];
19     int mmi[10];
20     node operator +(const node &t)const{
21         node tmp;
22         for(int i=0;i<9;i++){
23             tmp.mi[i]=min(mi[i],t.mi[i]);
24 
25             if(mi[i]<t.mi[i]){
26                 tmp.mmi[i]=min(mmi[i],t.mi[i]);
27             }
28             else{
29                 tmp.mmi[i]=min(mi[i],t.mmi[i]);
30             }
31         }
32         return tmp;
33     }
34 }a[N*4];
35 void up(int o,int l,int r,int k,int v){
36     if(l==r){
37         for(int i=0;i<9;i++){
38             a[o].mi[i]=a[o].mmi[i]=h;
39         }
40         int t=v;
41         for(int i=0;i<9;i++){
42             int x=t%10;
43             if(x){
44                 a[o].mi[i]=min(a[o].mi[i],v);
45             }
46             t/=10;
47         }
48         return;
49     }
50     int m=(l+r)/2;
51     if(k<=m)up(ls,l,m,k,v);
52     else up(rs,m+1,r,k,v);
53     a[o]=a[ls]+a[rs];
54 }
55 node qu(int o,int l,int r,int ql,int qr){
56     if(l>=ql&&r<=qr){
57         return a[o];
58     }
59     int m=(l+r)/2;
60     node res;
61     for(int i=0;i<9;i++){
62         res.mi[i]=res.mmi[i]=h;
63     }
64     if(ql<=m)res=res+qu(ls,l,m,ql,qr);
65     if(qr>m)res=res+qu(rs,m+1,r,ql,qr);
66     return res;
67 }
68 int main(){
69     scanf("%d%d",&n,&q);
70     for(int i=1;i<=n;i++){
71         int x;
72         scanf("%d",&x);
73         up(1,1,n,i,x);
74     }
75     while(q--){
76         int t,x,y;
77         scanf("%d%d%d",&t,&x,&y);
78         if(t==1)up(1,1,n,x,y);
79         else{
80             node res=qu(1,1,n,x,y);
81             ll ans=h;
82             for(int i=0;i<9;i++){
83                 ans=min(ans,1ll*res.mi[i]+res.mmi[i]);
84             }
85             if(ans==h)ans=-1;
86             printf("%d\n",ans);
87         }
88     }
89 }