HDU 4027 Can you answer these queries?【线段树区间更新】

Can you answer these queries?

Problem Description

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
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

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

Case #1:
19
7
6

题解

两种操作，一种是[l,r]间所有值开根号，在数据范围内一直开根号最后都会变成1，所有如果区间和为 l-r+1 就可以跳过

另一种就是求区间和

需注意的点： 输入的 l 可能比 r 小 需要swap一下 ；数据较大用 long long

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn =  1e5+10  ;
const int inf =	0x3f3f3f3f;

struct node{
    ll l,r;
    ll sum;
}tree[maxn<<2];

int kase=0;
int n,m,t;
int p,q;
int x,y;
ll a,b,c;
int val = 1;
ll ans = 0;
void pushup(int k)
{
    tree[k].sum = tree[k<<1].sum+tree[k<<1|1].sum;
}
void build(int l,int r,int k)
{
    tree[k].l = l;  tree[k].r = r;
    if(l == r){ scanf("%lld",&tree[k].sum); return ;  }
    int mid = (l+r)>>1;
    build(l,mid,k<<1);
    build(mid+1,r,k<<1|1);
    pushup(k);
}
void updata(int k)
{
    if(tree[k].l==tree[k].r)
    {
       tree[k].sum = sqrt(tree[k].sum );
       return ;
    }
    if(a <= tree[k].l && b >= tree[k].r && tree[k].sum == tree[k].r-tree[k].l+1)
    {
      return ;
    }
    
    int mid = (tree[k].l+tree[k].r)>>1;
    if(a<=mid){  updata(k<<1); }
    if(b>mid){  updata(k<<1|1);  }
    pushup(k);
}
void query(int k)
{
    if(a <= tree[k].l && b >= tree[k].r)
    {
      ans +=tree[k].sum ;
      return ;
    }
    int mid = (tree[k].l+tree[k].r)>>1;
    if(a <= mid){ query(k<<1);}
    if(b > mid){ query(k<<1|1);}
}
int main()
{
  while(scanf("%d",&n)!=EOF)
  {
    build(1,n,1);
    scanf("%d",&m);
    printf("Case #%d:\n",++kase);
    while(m--)
    {
      scanf("%d%lld%lld",&q,&a,&b);
      if(a>b) swap(a,b);
      if(q)
      {
        ans = 0;
        query(1);
        printf("%lld\n",ans);
      }
      else  updata(1);
    }
    printf("\n");
  }
}