SPOJ 2713(GSS4-线段树区间开方-多组数据)

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本文介绍了一种使用C++解决特定数据操作问题的方法，包括修改元素为正平方根并求和的操作，适用于大规模数据集。文章详细阐述了解决方案的实现步骤和关键点，包括数据初始化、构建数据结构、查询操作以及更新操作等。

2713. Can you answer these queries IV

Problem code: GSS4

You are given a sequence A of N(N <= 100,000) positive integers. There sum will be less than 10¹⁸. On this sequence you have to apply M (M <= 100,000) operations:

(A) For given x,y, for each elements between the x-th and the y-th ones (inclusively, counting from 1), modify it to its positive square root (rounded down to the nearest integer).

(B) For given x,y, query the sum of all the elements between the x-th and the y-th ones (inclusively, counting from 1) in the sequence.

Input

Multiple test cases, please proceed them one by one. Input terminates by EOF.

For each test case:

The first line contains an integer N. The following line contains N integers, representing the sequence A₁..A_N.
The third line contains an integer M. The next M lines contain the operations in the form "i x y".i=0 denotes the modify operation, i=1 denotes the query operation.

Output

For each test case:

Output the case number (counting from 1) in the first line of output. Then for each query, print an integer as the problem required.

Print an blank line after each test case.

See the sample output for more details.

Example

Input:
5
1 2 3 4 5
5
1 2 4
0 2 4
1 2 4
0 4 5
1 1 5
4
10 10 10 10
3
1 1 4
0 2 3
1 1 4

Output:
Case #1:
9
4
6

Case #2:
40
26

这题我在交了《花神游历各国》以后直接Wa了……

定睛一看：

1.多组数据

2.a[i]取值≤10^18

3.l>r

4.数组初始化会T……

于是我T_Y

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MAXN (300000+10)
#define MAXM (600000+10)
#define Lson (x<<1)
#define Rson ((x<<1)+1)
long long n,m,a[MAXN*2]={0},b[MAXN*2]={0};
long long sum[MAXN*2];
void update(int x)
{
	if (b[Lson]==2||b[Rson]==2) b[x]=2;
	else b[x]=1; 
	sum[x]=sum[Lson]+sum[Rson];
}
void build(int x,int l,int r)
{
	if (l==r) 
	{
		scanf("%lld",&a[x]);sum[x]=a[x];
		if (a[x]<=1) b[x]=1;
		else b[x]=2;
	}		
	if (l>=r) return;
	int m=(l+r)>>1;
	build(Lson,l,m);
	build(Rson,m+1,r);
	update(x);
}
void pushdown(int x,int l,int r)
{
	if (b[x]<=1) return;
	if (l==r)
	{
		a[x]=sum[x]=sqrt(a[x]);//floor(sqrt((double)a[x])+0.5);
		if (a[x]<=1) b[x]=1;
		return;
	}
	int m=(l+r)>>1;
	pushdown(Lson,l,m);
	if (m<r) pushdown(Rson,m+1,r);
	update(x);
}
void change(int x,int l,int r,int L,int R)
{
	if (b[x]<=1) return;
	int m=(l+r)>>1;
	if (L<=l&&r<=R)
	{
		pushdown(x,l,r);
		return;
	}
	if (L<=m) change(Lson,l,m,L,R);
	if (m<R) change(Rson,m+1,r,L,R);
	update(x);	
}
long long qur(int x,int l,int r,int L,int R)
{
	int m=(l+r)>>1;
	if (L<=l&&r<=R)
	{
		return sum[x];
	}
	long long ans=0;
	if (L<=m) ans+=qur(Lson,l,m,L,R);
	if (m<R) ans+=qur(Rson,m+1,r,L,R);
	return ans;	
}

int main()
{
//	freopen("spoj2713.in","r",stdin);
//	freopen(".out","w",stdout);
	int tt=1;
	while (scanf("%d",&n)!=-1)
	{
	//	memset(a,0,sizeof(a));
	//	memset(b,0,sizeof(b));
	//	memset(sum,0,sizeof(sum));
		printf("Case #%d:\n",tt++);
		build(1,1,n);
		scanf("%d",&m);
		For(i,m)
		{
			int x,l,r;
			scanf("%d%d%d",&x,&l,&r);
			if (l>r) swap(l,r);
			if (x==1) printf("%lld\n",qur(1,1,n,l,r));
			else change(1,1,n,l,r);
		}
		puts("");
	}
	return 0;
}