[BZOJ3211&3038][上帝造题的七分钟2&花神游历各国][线段树]

题目大意：

给定一棵线段树，要求支持区间开根和区间查询。

思路：

直接上线段树暴力修改就好咯，至于为什么，因为每个节点修改的次数是有限的，一个数开根多次后肯定会变成1，然后再执行开根操作的时候就直接跳过就好了（1的根号向下取整还是1）。

坑：

花神游历各国可能会出现0

代码：

BZOJ3211:

#include <bits/stdc++.h>using namespace std;const int Maxn = 200100;typedef long long ll;ll num[Maxn << 2], mm[Maxn << 2];ll n, m;const ll Max(const ll &a, const ll &b) {    return a > b ? a : b;}inline void read(ll &x) {    x = 0; static char c;    for (; !(c >= '0' && c <= '9'); c = getchar());    for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = getchar());}inline void build(ll o, ll l, ll r) {    if (l == r) {        read(num[o]);        mm[o] = num[o];        return ;    }    ll mid = (l + r) >> 1;    build(o << 1, l, mid);    build(o << 1 | 1, mid + 1, r);    num[o] = num[o << 1] + num[o << 1 | 1];    mm[o] = Max(mm[o << 1], mm[o << 1 | 1]);}inline void cover(ll o, ll l, ll r, ll L, ll R) {    if (mm[o] <= 1) return;    if (l == r && l >= L && r <= R) {        num[o] = sqrt(num[o] * 1.0);         mm[o] = num[o];        return;     }    ll mid = (l + r) >> 1;    if (L <= mid) cover(o << 1, l, mid, L, R);    if (mid < R) cover(o << 1 | 1, mid + 1, r, L, R);    num[o] = num[o << 1] + num[o << 1 | 1];    mm[o] = Max(mm[o << 1], mm[o << 1 | 1]);}inline ll query(ll o, ll l, ll r, ll L, ll R) {    if (l >= L && r <= R) {        return num[o];    }    ll mid = (l + r) >> 1;    ll ret = 0;    if (L <= mid) ret = ret + query(o << 1, l, mid, L, R);    if (mid < R) ret = ret + query(o << 1 | 1, mid + 1, r, L, R);    return ret;}int main(void) {    //freopen("in.txt", "r", stdin);    read(n);    //  prllf("Case #%d:\n", ++group);        build(1, 1, n);        read(m);        ll x, a, b;        while(m--) {            read(x); read(a), read(b);            if (a > b) swap(a, b);            if (x == 1) {                printf("%lld\n", query(1, 1, n, a, b));            } else {                cover(1, 1, n, a, b);            }        }        //putchar('\n');    return 0;}

BZOJ3038:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int Maxn = 200100;typedef unsigned long long ulld;struct Int {    ulld a[5];    Int(void) {        memset(a, 0, sizeof(a));    }    Int operator + (Int b) const {        Int ret;        for (int i = 0; i < 5; i++) {            ret.a[i] += a[i] + b.a[i];            ret.a[i + 1] += ret.a[i] / 100000000000;            ret.a[i] %= 100000000000;        }        return ret;    }    bool same(ulld num) {        if (a[1] || a[2] || a[3] || a[4]) return false;        return a[0] == num;    }    void value(ulld x) {        a[1] = a[2] = a[3] = a[4] = 0;        a[0] = x;    }    ulld show(void) {        return a[0] + a[1] * 100;    }    void out(void) {        if (!(a[1] || a[2] || a[3] || a[4] || a[0])) {            putchar('0');            return;        }        bool output = false;        for (int i = 4; i >= 0; i--) {            if (a[i]) {                if (output) {                    int size = 11 - log10(a[i]);                    for (int i = 1; i <= size; i++) putchar('0');                }                printf ("%llu", a[i]);                output = true;            }            else if (output) printf("00000000000");        }    }};Int num[Maxn << 2];int n;ulld m;inline void read(ulld &x) {    x = 0; static char c;    for (; !(c >= '0' && c <= '9'); c = getchar());    for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = getchar());}//inline void read(ulld &x) {//  scanf("%llu", &x);//}inline void build(int o, int l, int r) {    if (l == r) {        ulld tmp;        read(tmp); num[o].value(tmp);        return ;    }    int mid = (l + r) >> 1;    build(o << 1, l, mid);    build(o << 1 | 1, mid + 1, r);    num[o] = num[o << 1] + num[o << 1 | 1];}inline void cover(int o, int l, int r, int L, int R) {    if (num[o] .same((ulld)(r - l + 1))) return;    if (l == r && l >= L && r <= R) {        num[o].value((ulld) sqrt(num[o].show() * 1.0));         return;     }    int mid = (l + r) >> 1;    if (L <= mid) cover(o << 1, l, mid, L, R);    if (mid < R) cover(o << 1 | 1, mid + 1, r, L, R);    num[o] = num[o << 1] + num[o << 1 | 1];}inline Int query(int o, int l, int r, int L, int R) {    if (l >= L && r <= R) {        return num[o];    }    int mid = (l + r) >> 1;    Int ret;    if (L <= mid) ret = ret + query(o << 1, l, mid, L, R);    if (mid < R) ret = ret + query(o << 1 | 1, mid + 1, r, L, R);    return ret;}int main(void) {    int group = 0;    scanf("%d", &n);    //  printf("Case #%d:\n", ++group);        build(1, 1, n);        read(m); int i;        ulld x, a, b;        while(m--) {            read(x); read(a), read(b);            if (a > b) swap(a, b);            if (x) {                Int ret = query(1, 1, n, a, b);                ret.out(); putchar('\n');            } else {                cover(1, 1, n, a, b);            }        }        //putchar('\n');    return 0;}

完。

By g1n0st