B 线段包含关系
You are given a sequence a1, a2, …, an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.

Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.

Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Input
The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.

Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.

Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:

(2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders;
(3, 2), (4, 2), (3, 5), (4, 5) — touch one border;
(5, 2), (2, 5) — match exactly.

#include<iostream>
#include<algorithm>
using namespace std;
struct node{
    
	long long l;
	long long r;
	long long len;
};
node a[1040000];
bool cmp(node a,node b){
    
	if(a.l==b.l)
	return a.r>b.r;//排序最关键一步，必须保证在L递增的条件下大R先出现
	return a.l<b.l;
}
long long ma;
long long item,item1;
int main(){
    
	long long n;
	cin>>n;
	for(long long  i=0;i<n;i++){
    
		scanf("%lld%lld",&a[i].l,&a[i].r);
		a[i].len=i+1;
	}
	sort(a,a+n,cmp);
	for(long long  i=0;i<n;i++){
    
		if(a[i].r>ma){
              //当出现大于ma的r时则可以交换并记录位置
			item=a[i].len;
			ma=a[i].r;
		}
		else {
    
			cout<<a[i].len<<" "<<item;//但ma>r时排序时保证了L1<=L2最后符合条件输出
			item1=1; 
			break;
		}
	}
	if(item1==0)
	cout<<"-1"<<" "<<"-1";
	return 0;
} 

C题
You might have heard about the next game in Lara Croft series coming out this year. You also might have watched its trailer. Though you definitely missed the main idea about its plot, so let me lift the veil of secrecy.

Lara is going to explore yet another dangerous dungeon. Game designers decided to use good old 2D environment. The dungeon can be represented as a rectangle matrix of n rows and m columns. Cell (x, y) is the cell in the x-th row in the y-th column. Lara can move between the neighbouring by side cells in all four directions.

Moreover, she has even chosen the path for herself to avoid all the traps. She enters the dungeon in cell (1, 1), that is top left corner of the matrix. Then she goes down all the way to cell (n, 1) — the bottom left corner. Then she starts moving in the snake fashion — all the way to the right, one cell up, then to the left to the cell in 2-nd column, one cell up. She moves until she runs out of non-visited cells. n and m given are such that she always end up in cell (1, 2).

Lara has already moved to a neighbouring cell k times. Can you determine her current position?
Input
The only line contains three integers n, m and k (2 ≤ n, m ≤ 109, n is always even, 0 ≤ k < n·m). Note that k doesn’t fit into 32-bit integer type!

Output
Print the cell (the row and the column where the cell is situated) where Lara ends up after she moves k times.

Examples
Input
4 3 0
Output
1 1
Input
4 3 11
Output
1 2
Input
4 3 7
Output
3 2
Note
Here is her path on matrix 4 by 3:

#include<iostream>
using namespace std;
int main(){
    //找到规律和简单思路更容易编写代码 
	long long n,m,k;
	cin>>n>>m>>k;
	if(k>n*m-1)cout<<1<<" "<<2;
	else if(k<=n-1&&k)
	cout<<k+1<<" "<<1;
	else if(k==0)cout<<1<<" "<<1;
	else {
    
		k-=n-1;//关键步骤不要忘记 
		long long item=k%(m-1);
		k/=m-1;
		if(item==0){
    
			if(k%2==0){
    
				cout<<n-k+1<<" "<<2;
			}
			else cout<<n-k+1<<" "<<m;
		}
		else {
    
			if(k%2==0){
    
				cout<<n-k<<" "<<item+1;
			}
			else{
    
				cout<<n-k<<" "<<m-item+1;
			}
		}
	}
	return 0;
}

D
Recently Max has got himself into popular CCG “BrainStone”. As “BrainStone” is a pretty intellectual game, Max has to solve numerous hard problems during the gameplay. Here is one of them:

Max owns n creatures, i-th of them can be described with two numbers — its health hpi and its damage dmgi. Max also has two types of spells in stock:

Doubles health of the creature (hpi := hpi·2);
Assigns value of health of the creature to its damage (dmgi := hpi).
Spell of first type can be used no more than a times in total, of the second type — no more than b times in total. Spell can be used on a certain creature multiple times. Spells can be used in arbitrary order. It isn’t necessary to use all the spells.

Max is really busy preparing for his final exams, so he asks you to determine what is the maximal total damage of all creatures he can achieve if he uses spells in most optimal way.

Input
The first line contains three integers n, a, b (1 ≤ n ≤ 2·105, 0 ≤ a ≤ 20, 0 ≤ b ≤ 2·105) — the number of creatures, spells of the first type and spells of the second type, respectively.

The i-th of the next n lines contain two number hpi and dmgi (1 ≤ hpi, dmgi ≤ 109) — description of the i-th creature.

Output
Print single integer — maximum total damage creatures can deal.

Examples
Input
2 1 1
10 15
6 1
Output
27
Input
3 0 3
10 8
7 11
5 2
Output
26
Note
In the first example Max should use the spell of the first type on the second creature, then the spell of the second type on the same creature. Then total damage will be equal to 15 + 6·2 = 27.

In the second example Max should use the spell of the second type on the first creature, then the spell of the second type on the third creature. Total damage will be equal to 10 + 11 + 5 = 26.

//最难之处在于 想到存在多次使用a于一个生物 对于最优生物的筛选过程，要尽量优化算法 
#include<iostream>
#include<algorithm>
using namespace std;
struct node{
    
	long long h;
	long long d;
	long long sub;
}d[1000000];
bool cmp(node a,node b){
    
	return a.sub>b.sub;//因为此题只有需将一个生物进行生命加倍，所以要找出最优解，而排序的目的是将b操作做到最优（局部贪心）。 
}
int main(){
    
	long long n;
	long long a,b;
	long long ma=0;
	cin>>n>>a>>b;
	for(int i=0;i<n;i++){
    
		scanf("%lld%lld",&d[i].h,&d[i].d);
	    d[i].sub=d[i].h-d[i].d;
	    ma+=d[i].d;
	}
	sort(d,d+n,cmp);
	long long sum=0;
	for(int i=0;i<n;i++){
    
	    if(i<b){
    
	    	if(d[i].sub>=0)sum+=d[i].h;
	    	else sum+=d[i].d;
		}
		else sum+=d[i].d;
	}
	long long s=0;
	if(b>=1)
	for(int i=0;i<n;i++){
    
		if(i<b){
    
			if(d[i].sub>=0){
    
				s=sum+(d[i].h<<a)-d[i].h;
			}
			else s=sum-d[i].d+(d[i].h<<a);
		}
		else{
    
			if(i==b&&d[b-1].sub>0){
    //然后因为a操作是在b的基础上所以将最后一个d[b-1]进行去除b操作，然后对于其余元素进行筛选 
				sum-=d[b-1].h;
				sum+=d[b-1].d;
			}
			s=sum-d[i].d+(d[i].h<<a);
		}
		if(s>ma)ma=s;
	}
	cout<<ma<<endl;
	return 0;
}