思路：线段树维护矩阵乘法。

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
    
     using namespace std;#define y1 y11#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb emplace_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
     
      #define pli pair
      
       #define pii pair
       
        #define piii pair
        
         #define pdd pair
         
          #define mem(a, b) memset(a, b, sizeof(a))#define debug(x) cerr << #x << " = " << x << "\n";#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);//headconst int INF = 0x3f3f3f3f;const int N = 2e5 + 10;struct Matrix { int a[5][5]; inline void init() { memset(a, INF, sizeof a); } inline Matrix operator * (const Matrix & rhs) const { Matrix res; res.init(); for (int k = 0; k < 5; ++k) { for (int i = 0; i < 5; ++i) { for (int j = 0; j < 5; ++j) { res.a[i][j] = min(res.a[i][j], a[i][k]+rhs.a[k][j]); } } } return res; }}tree[N<<2];char s[N];int n, q, l, r;inline int push_up(int rt) { tree[rt] = tree[rt<<1]*tree[rt<<1|1];}/* |2|0|1|7| */void build(int rt, int l, int r) { if(l == r) { tree[rt].init(); for (int i = 0; i < 5; ++i) tree[rt].a[i][i] = 0; if(s[l] == '2') tree[rt].a[0][1] = 0, tree[rt].a[0][0] = 1; else if(s[l] == '0') tree[rt].a[1][2] = 0, tree[rt].a[1][1] = 1; else if(s[l] == '1') tree[rt].a[2][3] = 0, tree[rt].a[2][2] = 1; else if(s[l] == '7') tree[rt].a[3][4] = 0, tree[rt].a[3][3] = 1; else if(s[l] == '6') tree[rt].a[3][3] = 1, tree[rt].a[4][4] = 1; return ; } int m = l+r >> 1; build(ls); build(rs); push_up(rt);}Matrix query(int L, int R, int rt, int l, int r) { if(L <= l && r <= R) return tree[rt]; int m = l+r >> 1; Matrix res; res.init(); for (int i = 0; i < 5; ++i) res.a[i][i] = 0; if(L <= m) res = res * query(L, R, ls); if(R > m) res = res* query(L, R, rs); return res;}int main() { scanf("%d %d", &n, &q); scanf("%s", s+1); build(1, 1, n); while(q--) { scanf("%d %d", &l, &r); Matrix res = query(l, r, 1, 1, n); if(res.a[0][4] == INF) printf("-1\n"); else printf("%d\n", res.a[0][4]); } return 0;}