Can you answer these queries?（线段树）-优快云博客

本文介绍了一个关于连续区间内数值更新与查询的问题，通过使用线段树优化数据结构，实现对一连串数值进行平方根操作及求和查询，旨在解决战场上的武器攻击效果评估。

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http://acm.hust.edu.cn/vjudge/contest/view.action?cid=66989#problem/H

这一题要注意剪枝，所有不必访问的点都不需要再访问了，注意最小的值不是0而是1；

H - Can you answer these queries?

Time Limit:2000MS Memory Limit:65768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4027

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 2 ⁶³.
  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

       
        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

Sample Output

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <string>
#define SI(T)int T;scanf("%d",&T)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
const int SIZE=1e5+10;
const int maxn=1<<30;
__int64 sum[SIZE<<2];
void pushup(int rt){
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt){
    if(l==r){
        scanf("%I64d",&sum[rt]);
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
void update(int L,int R,int l,int r,int rt){
    if(sum[rt]==r-l+1)return;
    if(l==r){
        sum[rt]=(__int64)sqrt(sum[rt]);
        return;
    }
    int m=(l+r)>>1;
    if(L<=m)update(L,R,lson);
    if(R>m)update(L,R,rson);
    pushup(rt);
}
__int64 query(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        return sum[rt];
    }
    if(sum[rt]==1)return 1;
    int m=(l+r)>>1;
    __int64 ans=0;
    if(L<=m)ans+=query(L,R,lson);
    if(R>m)ans+=query(L,R,rson);
    return ans;
}
int main()
{
    int n,m,t,x,y,cas=1;
    while(scanf("%d",&n)!=EOF){
         printf("Case #%d:\n",cas++);
        build(1,n,1);
        scanf("%d",&m);
        while(m--){
            scanf("%d%d%d",&t,&x,&y);
            if(x>y)swap(x,y);
            if(t==0){
                update(x,y,1,n,1);
            }
            else printf("%I64d\n",query(x,y,1,n,1));
        }
        printf("\n");
    }
    return 0;
}