花神游历各国 bzoj3211 线段树

Description

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有一个n个数的序列A，请写一个程序完成下列操作：
1 l r表示查询l到r的和
2 l r表示把每个a[x] (l<=x<=r)变为sqrt(x)
n<=100000,m<=100000

Solution

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其实我是偶然看过幂伟的题解的(滑稽
109 最多开根6次就变成1了，对于1无论怎样开根都是不影响的

两种做法。如果是树状数组的话用链表或者并查集记录一下每个数字下一个不是1的数字的位置，这样子均摊就是接近O(1)的了

线段树呢记录一下当前区间是否全部为1，然后这样即使日到底也不惧了。实测线段树会慢一点，但是好写啊

权限题啊，什么时候我也能有bzoj的权限号呢？

Code

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#include <stdio.h>#include <math.h>#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)#define ll long long#define N 100001struct treeNode{int l, r; ll sum; int c;}t[N * 4 + 1];inline ll read(){    char ch = getchar(); ll x = 0;    while (ch < '0' || ch > '9'){        ch = getchar();    }    while (ch <= '9' && ch >= '0'){        x = (x << 1) + (x << 3) + ch - '0';        ch = getchar();    }    return x;}inline ll query(int now, int l, int r){    int mid = (t[now].l + t[now].r) >> 1;    if (t[now].l == l && t[now].r == r){        return t[now].sum;    }else if (r <= mid){        return query(now << 1, l, r);    }else if (l > mid){        return query(now << 1 | 1, l, r);    }else{        return query(now << 1, l, mid) + query(now << 1 | 1, mid + 1, r);    }}inline void modify(int now, int l, int r){    int mid = (t[now].l + t[now].r) >> 1;    if (t[now].l == l && t[now].r == r){        if (t[now].c == 1){            return;        }else if (l == r){            t[now].sum = sqrt(t[now].sum);            t[now].c = t[now].sum == 1;        }else{            modify(now << 1, l, mid);            modify(now << 1 | 1, mid + 1, r);            t[now].sum = t[now << 1].sum + t[now << 1 | 1].sum;            t[now].c = t[now << 1].c & t[now << 1 | 1].c;        }        return;    }    if (r <= mid){        modify(now << 1, l, r);    }else if (l > mid){        modify(now << 1 | 1, l, r);    }else{        modify(now << 1, l, mid);        modify(now << 1 | 1, mid + 1, r);    }    t[now].sum = t[now << 1].sum + t[now << 1 | 1].sum;    t[now].c = t[now << 1].c & t[now << 1 | 1].c;}inline void build(int now, int l, int r){    int mid = (l + r) >> 1;    t[now] = (treeNode){l, r, 0, 0};    if (l == r){        t[now].sum = read();        t[now].c = t[now].sum == 1;        return;    }    build(now << 1, l, mid);    build(now << 1 | 1, mid + 1, r);    t[now].sum = t[now << 1].sum + t[now << 1 | 1].sum;    t[now].c = t[now << 1].c & t[now << 1 | 1].c;}int main(void){    int n = read();    build(1, 1, n);    int T = read();    while (T --){        int opt = read(), l = read(), r = read();        if (opt == 1){            printf("%lld\n", query(1, l, r));        }else if (opt == 2){            modify(1, l, r);        }    }    return 0;}