Title: https: //nanti.jisuanke.com/t/41350
Analysis: The first string transposed over
State transition, because only five states, the state i to state j minimum cost to enumerate [k] [i] -> [j] [k] is the minimum value (0 <= k <= 4)
0: Initial state
1：2
2：20
3：201
4：2019
MAT [i] [j] represents the minimum cost of state i is transferred to the j
#include<bits/stdc++.h> using namespace std; #define lson root<<1,l,midd #define rson root<<1|1,midd+1,r const int N=2e5+5; struct node{ int mat[5][5]; void init(){ memset(mat,0x3f,sizeof(mat)); } node operator + (const node &b){ node ret; for(int i=0;i<5;i++) for(int j=0;j<5;j++){ ret.mat[i][j]=N; for(int k=0;k<5;k++) ret.mat[i][j]=min(ret.mat[i][j],mat[i][k]+b.mat[k][j]); } return ret; } }tree[N<<2],ANS; char s[N]; void build(int root,int l,int r){ if(l==r){ for(int i=0;i<5;i++) for(int j=0;j<5;j++) if(j!=i) tree[root].mat[i][j]=N; else tree[root].mat[i][j]=0; if(s[l]=='8') tree[root].mat[4][4]=1,tree[root].mat[3][3]=1; else if(s[l]=='9') tree[root].mat[3][3]=1,tree[root].mat[3][4]=0; else if(s[l]=='1') tree[root].mat[2][2]=1,tree[root].mat[2][3]=0; else if(s[l]=='0') tree[root].mat[1][1]=1,tree[root].mat[1][2]=0; else if(s[l]=='2') tree[root].mat[0][0]=1,tree[root].mat[0][1]=0; return ; } int midd=(l+r)>>1; build(lson); build(rson); tree[root]=tree[root<<1]+tree[root<<1|1]; } void query(int L,int R,int root,int l,int r){ if(L<=l&&r<=R){ ANS=ANS+tree[root]; return ; } int midd=(l+r)>>1; if(L<=midd) query(L,R,lson); if(R>midd) query(L,R,rson); } char f[N]; int main(){ int n,t; scanf("%d%d",&n,&t); scanf("%s",f+1); for(int i=1,j=n;i<=n;i++,j--) s[i]=f[j]; //cout<<endl; build(1,1,n); while(t--){ int l,r; scanf("%d%d",&l,&r); int L=n-r+1,R=n-l+1; ANS.init (); for ( int i = 0 ; i < 5 ; i ++ ) ANS.mat [i] [i] = 0 ; query (L, R, 1 , 1 , n); int years = ANS.mat [ 0 ] [ 4 ]; if (age == N) years = - 1 ; printf ( " % d \ n " , year); } Return 0 ; }