Can you answer these queries?
Time Limit: 1 Sec Memory Limit: 256 MB
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5195
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.
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
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.
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
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
HINT
题意
线段树 维护区间开根号，查询区间和
题解：
直接搞就好了，直接区间维护线段和，然后开根号，要注意的一点就是，int范围内的数，最多开7次根号
代码：
//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff; //нчоч╢С
const int inf=0x3f3f3f3f;
/* inline void P(int x)
{
Num=0;if(!x){putchar('0');puts("");return;}
while(x>0)CH[++Num]=x%10,x/=10;
while(Num)putchar(CH[Num--]+48);
puts("");
}
*/
//**************************************************************************************
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
inline void P(int x)
{
Num=;if(!x){putchar('');puts("");return;}
while(x>)CH[++Num]=x%,x/=;
while(Num)putchar(CH[Num--]+);
puts("");
} struct node
{
int l,r;
ll cnt,sum;
};
node a[maxn*];
ll d[maxn];
void build(int x,int l,int r)
{
a[x].l=l,a[x].r=r;
a[x].cnt=;
if(l==r)
a[x].sum=d[l];
else
{
int mid=(l+r)>>;
build(x<<,l,mid);
build(x<<|,mid+,r);
a[x].sum=a[x<<].sum+a[x<<|].sum;
}
}
void pushup(int x)
{
a[x].sum=a[x<<].sum+a[x<<|].sum;
if(a[x<<].cnt>=&&a[x<<|].cnt>=)
a[x].cnt=;
}
void update(int x,int st,int ed)
{
int l=a[x].l,r=a[x].r;
if(a[x].cnt>=)
{
a[x].sum=r-l+;
return;
}
if(st<=l&&r<=ed)
{
a[x].cnt+=;
if(l==r)
a[x].sum=sqrt(a[x].sum);
else
{
update(x<<,st,ed);
update(x<<|,st,ed);
pushup(x);
}
} else
{
int mid=(l+r)>>;
if(st<=mid) update(x<<,st,ed);
if(ed> mid) update(x<<|,st,ed);
pushup(x);
}
}
ll query(int x,int st,int ed)
{
int l=a[x].l,r=a[x].r;
if(st<=l&&r<=ed)
return a[x].sum;
else
{
int mid=(l+r)>>;
ll sum1=,sum2=;
if(st<=mid)
sum1=query(x<<,st,ed);
if(ed>mid)
sum2=query(x<<|,st,ed);
return sum1+sum2;
}
}
int n,m,f,b,c;
int main()
{ int cas=;
while(scanf("%d",&n)!=EOF)
{
memset(d,,sizeof(d));
memset(a,,sizeof(a));
for(int i=;i<=n;i++)
d[i]=read();
printf("Case #%d:\n",cas++);
scanf("%d",&m);
build(,,n);
for(int i=;i<m;i++)
{
f=read(),b=read(),c=read();
if(b>c)swap(b,c);
if(f==)update(,b,c);
else printf("%lld\n",query(,b,c));
}
puts("");
}
}