讲过无数次了。。很显然,一个$10^12$的数开方不超过$8$次后就会变为$1$
因此直接暴力更改即可,维护一下这段区间是否被全改为了$1$
双倍经验:https://www.luogu.org/problemnew/show/P4145
#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<cmath>
#define int long long
#define Pair pair<int, int>
#define fi first
#define se second
#define MP(x, y) make_pair(x, y)
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M;
int a[MAXN];
#define ls k << 1
#define rs k << 1 | 1
struct Node {
int l, r, w, f;
}T[MAXN];
void update(int k) {
if(T[ls].f && T[rs].f) T[k].f = 1;
T[k].w = T[ls].w + T[rs].w;
}
void Build(int k, int ll, int rr) {
T[k] = (Node) {ll, rr};
if(ll == rr) {
T[k].w = a[ll];
return ;
}
int mid = ll + rr >> 1;
Build(ls, ll, mid);
Build(rs, mid + 1, rr);
update(k);
}
void push(int k) {
if(T[k].f) return ;
if(T[k].l == T[k].r) {
T[k].w = sqrt(T[k].w);
if(T[k].w == 1) T[k].f = 1;
return ;
}
push(ls); push(rs);
update(k);
}
void IntSqrt(int k, int ll, int rr) {
if(ll <= T[k].l && T[k].r <= rr) {
if(T[k].f) return ;
push(k); return ;
}
int mid = T[k].l + T[k].r >> 1;
if(ll <= mid) IntSqrt(ls, ll, rr);
if(rr > mid) IntSqrt(rs, ll, rr);
update(k);
}
int IntSum(int k, int ll, int rr) {
if(ll <= T[k].l && T[k].r <= rr) return T[k].w;
int mid = T[k].l + T[k].r >> 1;
if(ll > mid) return IntSum(rs, ll, rr);
else if(rr <= mid) return IntSum(ls, ll, rr);
else return IntSum(ls, ll, rr) + IntSum(rs, ll, rr);
}
main() {
int tot = 0;
while(scanf("%d", &N) != EOF) {
printf("Case #%d:\n", ++tot);
for(int i = 1; i <= N; i++) a[i] = read();
Build(1, 1, N);
M = read();
while(M--) {
int k = read(), l = read(), r = read();
if(l > r) swap(l, r);
if(k == 0) IntSqrt(1, l, r);
else printf("%lld\n", IntSum(1, l, r));
}
puts("");
}
return 0;
}
/*
*/