题面

Sol

暴力开根，一个数开根到小于等于\(1\)就不用管了，维护区间\(max\)，\(max<=1\)就不用管这个区间，线段树

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(1e5 + 5);IL int Input(){    RG int x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}int n, m;ll mx[_ << 2], sum[_ << 2];IL void Build(RG int x, RG int l, RG int r){    if(l == r){        mx[x] = sum[x] = Input();        return;    }    RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;    Build(ls, l, mid), Build(rs, mid + 1, r);    mx[x] = max(mx[ls], mx[rs]), sum[x] = sum[ls] + sum[rs];}IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R){    if(mx[x] <= 1) return;    if(l == r){        sum[x] = mx[x] = sqrt(mx[x]);        return;    }    RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;    if(L <= mid) Modify(ls, l, mid, L, R);    if(R > mid) Modify(rs, mid + 1, r, L, R);    mx[x] = max(mx[ls], mx[rs]), sum[x] = sum[ls] + sum[rs];}IL ll Query(RG int x, RG int l, RG int r, RG int L, RG int R){    if(L <= l && R >= r) return sum[x];    RG int mid = (l + r) >> 1; RG ll ret = 0;    if(L <= mid) ret = Query(x << 1, l, mid, L, R);    if(R > mid) ret += Query(x << 1 | 1, mid + 1, r, L, R);    return ret;}int main(RG int argc, RG char* argv[]){    n = Input(), Build(1, 1, n), m = Input();    for(RG int i = 1; i <= m; ++i){        RG int k = Input(), l = Input(), r = Input();        if(k == 1) printf("%lld\n", Query(1, 1, n, l, r));        else Modify(1, 1, n, l, r);    }    return 0;}