hdu 4027 Can you answer these queries? 线段树 懒惰标记 单点更新妙用

题意：把一组舰队看成是线段上的端点，有一种秘密武器，每次可以攻击一个区间上的船，然后他们的防御力x减低为sqrt(x).

做法：最多只能有7次攻击有效。所以用带点更新吧，一遇到区间防御力总和为len(区间中含有舰船的个数)，就停止更新，因为这个区间所有的船的防御力已经只剩1了,这么一来，最后的时间复杂度不会太高

#include <cstdio>#include <cstring>#include <cmath>#define max(a, b) ((a) > (b) ? (a) : (b))#define left l, m, x << 1#define right m + 1, r, x << 1 | 1//有效攻最多七下。。。typedef __int64 LL;const int LMT = 100003;LL sum[LMT << 2];struct __node{    int l, r, len;}node[LMT << 2];inline LL get(void){    LL res = 0;    char ch = getchar();    while(ch < '0' || ch > '9') ch = getchar();    while(ch >= '0' && ch <= '9')    {        res = res * 10 + ch - '0';        ch = getchar();    }    return res;}void build(int l, int r, int x){    node[x].l = l;    node[x].r = r;    node[x].len = r - l + 1;    if(l == r)    {        sum[x] = get();        return ;    }    int m = (l + r) >> 1;    build(left);    build(right);    sum[x] = sum[x << 1] + sum[x << 1 | 1];}LL query(int L, int R, int x){    if (L <= node[x].l && node[x].r <= R) return sum[x];    int m = (node[x].l + node[x].r) >> 1;    LL res = 0;    if (L <= m) res += query(L, R, x << 1);    if (R > m) res += query(L, R, x << 1 | 1);    return res;}void update(int L, int R, int x){    if(sum[x] == node[x].len) return;    if(node[x].l == node[x].r)    {        sum[x] = (LL)sqrt(1.0 * sum[x]);        return;    }    int m = (node[x].l + node[x].r) >> 1;    if (L <= m) update(L, R, x << 1);    if (R > m) update(L, R, x << 1 | 1);        sum[x] = sum[x << 1] + sum[x << 1 | 1];}int main(void){    int n, q, ord, l, r, I = 1;    while(~scanf("%d", &n))    {        build(1, n, 1);        scanf("%d", &q);        printf("Case #%d:\n", I++);        while(q--)        {            scanf("%d%d%d", &ord, &l, &r);            if(l > r)            {             l = l ^ r;             r = l ^ r;             l = l ^ r;            }            if(ord) printf("%I64d\n", query(l, r, 1));            else update(l, r, 1);        }        printf("\n");    }    return 0;}