線段樹，樹狀數組，lazy優化
好睏不多bb了
一：
C - Ultra-QuickSort
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.
Input
The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
Output
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
Sample Input
5
9
1
0
5
4
3
1
2
3
0
Sample Output
6
0
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<memory.h>
using namespace std;
typedef long long ll;
int n;
struct node
{
int num,ci;
}a[500005];
int tree[500005];
int lowbit(int x)
{
return x&(-x);
}
void change(int p,int x)
{
for(int i=p;i<=n;i+=lowbit(i))
tree[i]+=x;
}
int query(int p)
{
int sum=0;
for(int i=p;i>0;i-=lowbit(i))
sum+=tree[i];
return sum;
}
bool cmp(node a,node b)
{
return a.num<b.num;
}
int main()
{
while(~scanf("%d",&n)&&n)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i].num);
a[i].ci=i;
change(i,1);
}
sort(a+1,a+1+n,cmp);
ll ans=0;
for(int i=1;i<=n;i++)
{
ans+=query(a[i].ci-1);
change(a[i].ci,-1);
}
printf("%lld\n",ans);
}
return 0;
}
二：
D - Can you answer these queries?
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
Case #1:
19
7
6
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<memory.h>
#include<math.h>
using namespace std;
typedef long long ll;
int n,m;
ll a[100005];
ll sum;
struct node
{
int l,r;
ll sum;
}b[100005*4];
void build(int l,int r,int p)
{
int mid;
b[p].l=l;
b[p].r=r;
if(l==r)
{
b[p].sum=a[l];
return;
}
mid=(l+r)/2;
build(l,mid,p*2);
build(mid+1,r,p*2+1);
b[p].sum=b[p*2].sum+b[p*2+1].sum;
}
void update2(int x,int p)
{
int mid;
if(b[p].l==b[p].r)
{
b[p].sum=a[b[p].l]=(int)(sqrt(1.0*a[b[p].l]));
return;
}
mid=(b[p].l+b[p].r)/2;
if(x>mid)update2(x,p*2+1);
else update2(x,p*2);
b[p].sum=b[p*2].sum+b[p*2+1].sum;
}
void update(int l,int r,int p)
{
int mid;
if(b[p].l==l&&b[p].r==r)
{
if(b[p].sum==b[p].r-b[p].l+1)return;
for(int j=b[p].l;j<=b[p].r;j++)
{
if(a[j]==1)continue;
update2(j,1);
}
return ;
}
mid=(b[p].l+b[p].r)/2;
if(r<=mid)update(l,r,p*2);
else if(l>mid)update(l,r,p*2+1);
else
{
update(l,mid,p*2);
update(mid+1,r,p*2+1);
}
}
void query(int l,int r,int p)
{
int mid;
if(b[p].l==l&&b[p].r==r)
{
sum+=b[p].sum;
return;
}
mid=(b[p].l+b[p].r)/2;
if(r<=mid)query(l,r,p*2);
else if(l>mid)query(l,r,p*2+1);
else
{
query(l,mid,p*2);
query(mid+1,r,p*2+1);
}
}
int main()
{
int cnt=1;
while(~scanf("%d",&n))
{
printf("Case #%d:\n",cnt++);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
build(1,n,1);
cin>>m;
for(int i=1;i<=m;i++)
{
int t,x,y;
scanf("%d%d%d",&t,&x,&y);
if(y<x)
{
int tmp;
tmp=x;
x=y;
y=tmp;
}
if(t==0)
{
update(x,y,1);
}
else
{
sum=0;
query(x,y,1);
printf("%lld\n",sum);
}
}
printf("\n");
}
return 0;
}
三：
E - Interesting Array
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value 
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=1e5+5;
int sum[maxn<<2],lazy[maxn<<2];
int l[maxn],r[maxn],c[maxn];
int now;
void pushdown(int i)
{
if(lazy[i])
{
sum[2*i]|=lazy[i];
sum[2*i+1]|=lazy[i];
lazy[2*i]|=lazy[i];
lazy[2*i+1]|=lazy[i];
lazy[i]=0;
}
}
void update(int i,int l,int r,int x,int y,int c)
{
if(x<=l&&r<=y)
{
sum[i]|=c;
lazy[i]|=c;
return;
}
pushdown(i);
int mid=(l+r)/2;
if(x<=mid) update(2*i,l,mid,x,y,c);
if(y>mid) update(2*i+1,mid+1,r,x,y,c);
sum[i]=sum[2*i]&sum[2*i+1];
return;
}
void query(int i,int l,int r,int x,int y)
{
if(x<=l&&r<=y)
{
now&=sum[i];
return;
}
pushdown(i);
int mid=(l+r)/2;
if(x<=mid) query(2*i,l,mid,x,y);
if(y>mid) query(2*i+1,mid+1,r,x,y);
sum[i]=sum[2*i]&sum[2*i+1];
return;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&l[i],&r[i],&c[i]);
update(1,1,n,l[i],r[i],c[i]);
}
for(int i=1;i<=m;i++)
{
now=(1<<30)-1;
query(1,1,n,l[i],r[i]);
if(now!=c[i])
{
cout<<"NO"<<endl;
return 0;
}
}
cout<<"YES"<<endl;
for(int i=1;i<=n;i++)
{
now=(1<<30)-1;
query(1,1,n,i,i);
cout<<now<<" ";
}
}
return 0;
}