GSS4 - Can you answer these queries IV || luogu4145上帝造题的七分钟2 / 花神游历各国 (线段树)

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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#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 <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#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
*/