[HDU] 4027

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 20101    Accepted Submission(s): 4737

Problem Description

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.

 

Input

The input contains several test cases, terminated by EOF.
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2⁶³.
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.

 

Output

For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.

 

Sample Input

101 2 3 4 5 6 7 8 9 1050 1 101 1 101 1 50 5 81 4 8

 

Sample Output

Case #1:1976

 

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

今天航神 小太阳和毛子一起去打 的比赛

有一道这个

他们没我做不出来啊哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈哈（假装只有我做得出来）

区间修改  每个元素开根号向下取整

区间查询求和

数据范围RT

恩一个数字最多根号 6 次就变成 1 了就没必要开根号

所以放心大胆的把区间更新变成单点更新

用线段树去维护

然后线段树维护一下每个节点包含的区间是不是全都是 1 

如果是就不用往下更新了

别的照常

这题输入数据是 LL 为啥用 int 会被 T 啊

HDU的评测姬这么傲娇的么

好像之前没写过区间修改换单点修改

刚开始有点写坏了

不过还是A了

#include <bits/stdc++.h>#define LL long longusing namespace std;const int N = 100010;struct Tree{    int l;    int r;    LL sum;    bool o;};Tree tree[N * 4];LL ma[N];int n, m;void up(int x){    if(tree[x << 1].o && tree[x << 1 | 1].o){        tree[x].o = true;    }    tree[x].sum = tree[x << 1].sum + tree[x << 1 | 1].sum;}void build(int x, int l, int r){    tree[x].l = l;    tree[x].r = r;    tree[x].o = false;    tree[x].sum = 0;    if(l == r){        tree[x].sum = ma[l];        if(tree[x].sum == 1){            tree[x].o = true;        }    }    else{        int mid;        mid = l + ((r - l) >> 1);        build(x << 1, l, mid);        build(x << 1 | 1, mid + 1, r);        up(x);    }}void update(int x, int l, int r){    int ll = tree[x].l;    int rr = tree[x].r;    if(tree[x].o || l > r){        return;    }    if(ll == rr){        tree[x].sum = (int)sqrt(tree[x].sum);        if(tree[x].sum == 1){            tree[x].o = true;        }    }    else{        int mid;        mid = ll + ((rr - ll) >> 1);        update(x << 1, l, min(mid, r));        update(x << 1 | 1, max(mid + 1, l), r);        up(x);    }}LL quary(int x, int l, int r){    int ll = tree[x].l;    int rr = tree[x].r;    if(l > r){        return 0;    }    if(l == ll && rr == r){        return tree[x].sum;    }    else{        int mid;        LL ans = 0;        mid = ll + ((rr - ll) >> 1);        ans += quary(x << 1, l, min(mid, r));        ans += quary(x << 1 | 1, max(mid + 1, l), r);        return ans;    }}int main(int argc, char const *argv[]){    int nc = 1;    while(scanf("%d", &n) == 1){        printf("Case #%d:\n", nc ++);        for(int i = 1; i <= n; i ++){            scanf("%lld", &ma[i]);        }        build(1, 1, n);        scanf("%d", &m);        for(int i = 0; i < m; i ++){            int t, l, r;            scanf("%d%d%d", &t, &l, &r);            if(l > r){                swap(l, r);            }            if(t){                printf("%lld\n", quary(1, l, r));            }            else{                update(1, l, r);            }        }        cout << endl;    }    return 0;}