GSS4 - Can you answer these queries IV || luogu4145上帝造题的七分钟2 / 花神游历各国
GSS4 - Can you answer these queries IV
题目链接:https://www.luogu.org/problemnew/show/SP2713
线段树经典题目,然而被我用分块A了.
对于区间开根号,\(1e18\)最多会被开\(6\)次就会成为\(1\),成为\(1\)后,再开根号也是\(1\),0开根号也是0,线段树(分块)维护区间所有的数是否全部小于等于1,如果不是,就暴力更新,如果是,那就不要更新这个区间.
时间复杂度\(O(\sqrt n * n)\)
分块CODE:
#include
#include
#include
#include
#include
#define ll long long
const int maxN = 100000 + 7;
ll a[maxN];
bool is_sqrt[maxN];
int L[maxN],R[maxN],belong[maxN];
ll sum[maxN];
int num[maxN];
inline ll read() {
ll x = 0,f = 1;char c = getchar();
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;
}
ll Query(int l,int r) {
int b_l = belong[l],b_r = belong[r];
ll ans = 0;
if(b_l == b_r) {
for(int i = l;i <= r;++ i)
ans += a[i];
return ans;
}
for(int i = b_l + 1;i < b_r;++ i) ans += sum[i];
for(int i = l;i <= R[b_l];++ i) ans += a[i];
for(int i = L[b_r];i <= r;++ i) ans += a[i];
return ans;
}
void Inter_sqrt(int l,int r) {
int b_l = belong[l],b_r = belong[r];
if(b_l == b_r) {
if(is_sqrt[b_l]) return;
for(int i = l;i <= r;++ i) {
if(a[i] == 1) continue;
sum[b_l] -=a[i];
a[i] = sqrt(a[i]);
sum[b_l] += a[i];
if(a[i] == 1) num[b_l] ++;
}
if(num[b_l] == R[b_l] - L[b_l] + 1) is_sqrt[b_l] = true;
return;
}
for(int i = b_l + 1;i < b_r;++ i) {
if(is_sqrt[i]) continue;
for(int j = L[i];j <= R[i];++ j) {
if(a[j] == 1) continue;
sum[i] -=a[j];
a[j] = sqrt(a[j]);
sum[i] += a[j];
if(a[j] == 1) num[i] ++;
}
if(num[i] == R[i] - L[i] + 1) is_sqrt[i] = true;
}
for(int i = l;i <= R[b_l];++ i) {
if(a[i] == 1) continue;
sum[b_l] -=a[i];
a[i] = sqrt(a[i]);
sum[b_l] += a[i];
if(a[i] == 1) num[b_l] ++;
}
if(num[b_l] == R[b_l] - L[b_l] + 1) is_sqrt[b_l] = true;
for(int i = L[b_r];i <= r;++ i) {
if(a[i] == 1) continue;
sum[b_r] -=a[i];
a[i] = sqrt(a[i]);
sum[b_r] += a[i];
if(a[i] == 1) num[b_r] ++;
}
if(num[b_r] == R[b_r] - L[b_r] + 1) is_sqrt[b_r] = true;
}
int main() {
int tot = 0;
int n;
while(scanf("%d",&n) == 1) {
printf("Case #%d:\n",++ tot);
memset(num,0,sizeof(num));
memset(L,0,sizeof(L));
memset(R,0,sizeof(R));
memset(sum,0,sizeof(sum));
memset(is_sqrt,0,sizeof(is_sqrt));
int q = sqrt(n);
for(int i = 1;i <= n;++ i)
a[i] = read();
for(int i = 1;i <= n;++ i){
belong[i] = i / q + 1;
sum[belong[i]] += a[i];
if(a[i] == 1) num[belong[i]] ++;
}
for(int i = 1;i <= n;++ i) R[belong[i]] = i;
for(int i = n;i >= 1;-- i) L[belong[i]] = i;
int m = read();
int opt,l,r;
while(m --) {
opt = read();l = read();r = read();
if(l > r) std::swap(l,r);
if(opt) printf("%lld\n",Query(l,r));
else Inter_sqrt(l,r);
}
}
return 0;
} luogu4145上帝造题的七分钟2 / 花神游历各国
用分块很难水过,我们用线段树维护区间最大值即可.
线段树CODE:
#include
#include
#include
#include
#define max(a,b) a > b ? a : b
#define ll long long
const ll maxN = 100000 + 7;
using namespace std;
struct Node{
ll l,r;
ll sum;
ll maxx;
}tree[maxN << 2];
ll a[maxN];
void swap(ll &a,ll &b) {
ll k = b;
b = a;
a = k;
}
inline ll read() {
ll x = 0,f = 1;char c = getchar();
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;
}
void updata(ll now) {
tree[now].sum = tree[now << 1].sum + tree[now << 1 | 1].sum;
tree[now].maxx = max(tree[now << 1 | 1].maxx,tree[now << 1].maxx);
}
void build(ll l,ll r,ll now) {
tree[now].l = l;
tree[now].r = r;
if(l == r) {
tree[now].sum = a[l];
tree[now].maxx = a[l];
return ;
}
ll mid = (l + r) >> 1;
build(l,mid,now << 1);
build(mid + 1,r,now << 1 | 1);
updata(now);
}
ll Query(ll l,ll r,ll now) {
if(tree[now].l >= l && tree[now].r <= r) return tree[now].sum;
ll mid = (tree[now].l + tree[now].r) >> 1;
ll sum = 0;
if(l <= mid) sum += Query(l,r,now << 1);
if(r > mid) sum += Query(l,r,now << 1 | 1);
return sum;
}
void work(ll now) {
if(tree[now].l == tree[now].r) {
ll L = tree[now].l;
a[L] = sqrt(a[L]);
tree[now].sum = a[L];
tree[now].maxx = a[L];
return;
}
if(tree[now << 1].maxx > 1) work(now << 1);
if(tree[now << 1 | 1].maxx > 1) work(now << 1 | 1);
updata(now);
return ;
}
void Inter_sqrt(ll l,ll r,ll now) {
if(tree[now].l >= l && tree[now].r <= r) {
if( tree[now].maxx > 1 ) work(now);
return ;
}
ll mid = (tree[now].l + tree[now].r) >> 1;
if(l <= mid) Inter_sqrt(l,r,now << 1);
if(r > mid) Inter_sqrt(l,r,now << 1 | 1);
updata(now);
return;
}
int main() {
ll n = read();
for(ll i = 1;i <= n;++ i)
a[i] = read();
build(1,n,1);
ll m = read();
ll opt,l,r;
while(m --) {
opt = read();l = read();r = read();
if(l > r)swap(l,r);
if(opt == 1) printf("%lld\n",Query(l,r,1));
else Inter_sqrt(l,r,1);
}
return 0;
}
/*
10
1 2 3 4 5 6 7 8 9 10
5
0 1 10
1 1 10
1 1 5
0 5 8
1 4 8
*/