E：01 串
分析：
线段树维护转移矩阵。每个节点是一个矩阵，区间内的矩阵乘起来就是答案矩阵。矩阵乘法满足结合律，所以线段树维护。
代码：
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 #include<cctype> 7 #include<set> 8 #include<queue> 9 #include<vector> 10 #include<map> 11 #define Root 1, n, 1 12 #define lson l, mid, rt << 1 13 #define rson mid + 1, r, rt << 1 | 1 14 using namespace std; 15 typedef long long LL; 16 17 inline int read() { 18 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 19 for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 20 } 21 22 const int N = 300005; 23 char s[N]; 24 struct Node{ 25 int a[2][2]; 26 void Calc(int v) { 27 if (v) a[0][0] = 1, a[1][0] = 1, a[0][1] = 2, a[1][1] = 0; 28 else a[0][0] = 0, a[1][0] = 2, a[0][1] = 1, a[1][1] = 1; 29 } 30 }T[N << 2]; 31 Node operator + (const Node &A,const Node &B) { 32 Node C; memset(C.a, 0x3f, sizeof(C.a)); 33 for (int k = 0; k < 2; ++k) 34 for (int i = 0; i < 2; ++i) 35 for (int j = 0; j < 2; ++j) 36 C.a[i][j] = min(C.a[i][j], A.a[i][k] + B.a[k][j]); 37 return C; 38 } 39 void build(int l,int r,int rt) { 40 if (l == r) { T[rt].Calc(s[l] - '0'); return ; } 41 int mid = (l + r) >> 1; 42 build(lson); build(rson); 43 T[rt] = T[rt << 1] + T[rt << 1 | 1]; 44 } 45 void update(int l,int r,int rt,int p,int v) { 46 if (l == r) { T[rt].Calc(v); return ; } 47 int mid = (l + r) >> 1; 48 if (p <= mid) update(lson, p, v); 49 else update(rson, p, v); 50 T[rt] = T[rt << 1] + T[rt << 1 | 1]; 51 } 52 Node query(int l,int r,int rt,int L,int R) { 53 if (L <= l && r <= R) return T[rt]; 54 int mid = (l + r) >> 1; 55 if (R <= mid) return query(lson, L, R); 56 else if (L > mid) return query(rson, L, R); 57 else return query(lson, L, R) + query(rson, L, R); 58 } 59 int main() { 60 int n = read(); 61 scanf("%s", s + 1); 62 build(Root); 63 Node ans; 64 for (int Q = read(); Q--; ) { 65 int opt = read(), x = read(), y = read(); 66 if (opt == 1) { 67 ans = query(Root, x, y); 68 printf("%d ",min(ans.a[0][0], ans.a[1][0] + 1)); 69 } 70 else update(Root, x, y); 71 } 72 return 0; 73 }