Codeforece Educational Codeforces Round 34 (Rated for Div. 2) (A-D)题解

A：

A. Hungry Student Problem

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Ivan's classes at the university have just finished, and now he wants to go to the local CFK cafe and eat some fried chicken.

CFK sells chicken chunks in small and large portions. A small portion contains 3 chunks; a large one — 7 chunks.
Ivan wants to eat exactly x chunks. Now he wonders whether he can buy exactly this amount of chicken.

Formally, Ivan wants to know if he can choose two non-negative integers a and b in
such a way that a small portions and b large
ones contain exactly x chunks.

Help Ivan to answer this question for several values of x!

Input

The first line contains one integer n (1 ≤ n ≤ 100)
— the number of testcases.

The i-th of the following n lines
contains one integer xi (1 ≤ xi ≤ 100)
— the number of chicken chunks Ivan wants to eat.

Output

Print n lines, in i-th
line output YES if Ivan can buy exactly xi chunks.
Otherwise, print NO.

Example

input

2
6
5

output

YES
NO

Note

In the first example Ivan can buy two small portions.

In the second example Ivan cannot buy exactly 5 chunks, since one small portion is not enough, but two small portions or one large is too much.

之前CF出过一个类似的题目：  3 7 组成的不能够凑成的数最大是 3*7 - (3+7)= 11  ,11 之后的一定可以凑出来；

  a,b 互质， a b. 最大不能凑成的数 为 a*b - (a+b)  

刚开始没看出来，  脑回路啊！！！

map<int,int>mp;
void init()
{
for(int i=0;i<=50;i++)
{
for(int j=0;j<=50;j++)
{
int x=i*3;
mp[x]=1;
int y=j*7;
mp[y]=1;
int z=x+y;
mp[z]=1;
}
}
}

int main()
{

int n,x;
init();
mp[0]=0;

cin>>n;
while(n--)
{
scanf("%d",&x)；
if(mp[x])
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;

}
return 0;
}

B;

B. The Modcrab

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vova is again playing some computer game, now an RPG. In the game Vova's character received a quest: to slay the fearsome monster called Modcrab.

After two hours of playing the game Vova has tracked the monster and analyzed its tactics. The Modcrab has h2 health
points and an attack power of a2.
Knowing that, Vova has decided to buy a lot of strong healing potions and to prepare for battle.

Vova's character has h1 health
points and an attack power of a1.
Also he has a large supply of healing potions, each of which increases his current amount of health points by c1 when
Vova drinks a potion. All potions are identical to each other. It is guaranteed that c1 > a2.

The battle consists of multiple phases. In the beginning of each phase, Vova can either attack the monster (thus reducing its health by a1)
or drink a healing potion (it increases Vova's health by c1; Vova's
health can exceed h1).
Then, if the battle is not over yet, the Modcrab attacks Vova, reducing his health by a2.
The battle ends when Vova's (or Modcrab's) health drops to 0 or lower. It is possible that the battle ends in a middle of a phase after Vova's
attack.

Of course, Vova wants to win the fight. But also he wants to do it as fast as possible. So he wants to make up a strategy that will allow him to win the fight after the minimum possible number of phases.

Help Vova to make up a strategy! You may assume that Vova never runs out of healing potions, and that he can always win.

Input

The first line contains three integers h1, a1, c1 (1 ≤ h1, a1 ≤ 100, 2 ≤ c1 ≤ 100)
— Vova's health, Vova's attack power and the healing power of a potion.

The second line contains two integers h2, a2 (1 ≤ h2 ≤ 100, 115183
 ≤ a2 < c1)
— the Modcrab's health and his attack power.

Output

In the first line print one integer n denoting the minimum number of phases required to win the battle.

Then print n lines. i-th
line must be equal to HEAL if Vova drinks a potion in i-th
phase, or STRIKE if he attacks the Modcrab.

The strategy must be valid: Vova's character must not be defeated before slaying the Modcrab, and the monster's health must be 0 or lower after
Vova's last action.

If there are multiple optimal solutions, print any of them.

Examples

input

10 6 100
17 5

output

4
STRIKE
HEAL
STRIKE
STRIKE

input

11 6 100
12 5

output

2
STRIKE
STRIKE

Note

In the first example Vova's character must heal before or after his first attack. Otherwise his health will drop to zero in 2 phases while he
needs 3 strikes to win.

In the second example no healing needed, two strikes are enough to get monster to zero health and win with 6 health left.

模拟可以过， 但是要特判， 当最后一次 h2 < a1   h1<a2 时 是可以直接打死不用加血的；

并且决策时 必须是二选一，

int main()
{
int h1,a1,c1;
int h2,a2;

vector<int>Q;
Q.clear();
scanf("%d %d %d",&h1,&a1,&c1);
scanf("%d %d",&h2,&a2);
int ans=0;
while(h2>0)
{
if(h1>a2)
{
ans++;
Q.push_back(0);
h2-=a1;
}
else //if(h1<=a2)
{
if(h2<=a1)
{
ans++;
Q.push_back(0);
h2-=a1;
}
else
{

ans++;
Q.push_back(1);
h1+=c1;
}

}
h1-=a2;
}

cout<<ans<<endl;
for(int i=0;i<Q.size();i++)
{
int t=Q[i];
if(t)
cout<<"HEAL"<<endl;
else
cout<<"STRIKE"<<endl;
}

return 0;
}

C:

C. Boxes Packing

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th
box is a cube with side length ai.

Mishka can put a box i into another box j if
the following conditions are met:

i-th box is not put into another box;

j-th box doesn't contain any other boxes;

box i is smaller than box j (ai < aj).

Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff
it is not put into some another box.

Help Mishka to determine the minimum possible number of visible boxes!

Input

The first line contains one integer n (1 ≤ n ≤ 5000)
— the number of boxes Mishka has got.

The second line contains n integers a1, a2,
..., an (1 ≤ ai ≤ 109),
where ai is
the side length of i-th box.

Output

Print the minimum possible number of visible boxes.

Examples

input

3
1 2 3

output

1

input

4
4 2 4 3

output

2

Note

In the first example it is possible to put box 1 into box 2,
and 2 into 3.

In the second example Mishka can put box 2 into box 3,
and box 4 into box 1.

C要比B 简单，   C题直接找 数目最多的就行了，  小的一定能够放在大的里（前提是一个大的只能放一个小的）

map<ll,ll>mp;
ll a[MAXN];
int main()
{
int n;
scanf("%d",&n);
ll ans=-INF;
for(int i=1;i<=n;i++)
{
scanf("%I64d",&a[i]);
mp[a[i]]++;
ans=max(ans,mp[a[i]]);
}
cout<<ans<<endl;
return 0;
}

D;

D. Almost Difference

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Let's denote a function



You are given an array a consisting of n integers.
You have to calculate the sum of d(ai, aj) over
all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Input

The first line contains one integer n (1 ≤ n ≤ 200000)
— the number of elements in a.

The second line contains n integers a1, a2,
..., an (1 ≤ ai ≤ 109)
— elements of the array.

Output

Print one integer — the sum of d(ai, aj) over
all pairs (i, j) such that 1 ≤ i ≤ j ≤ n.

Examples

input

5
1 2 3 1 3

output

4

input

46 6 5 5

output

0

input

46 6 4 4

output

-8

Note

In the first example:

d(a1, a2) = 0;

d(a1, a3) = 2;

d(a1, a4) = 0;

d(a1, a5) = 2;

d(a2, a3) = 0;

d(a2, a4) = 0;

d(a2, a5) = 0;

d(a3, a4) =  - 2;

d(a3, a5) = 0;

d(a4, a5) = 2.

一层for 循环 求  当前数于谦的差值， 减去差值为1 的；

用矩阵面积法求差值；   a[i]* (i-1) - sum(前i-1 项之和）

用map 数组标记，  小一的是差值为1  要减去，  大一的 另一半

long long 会爆- -   一个方法是 用long double  一个是用大数封装；

typedef long double ld;
map<ld,ld>mp;
ld a[MAXN];

int main()
{
ll n;
scanf("%I64d",&n);
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
ld sum=0;
ld ans=0;
for(int i=1;i<=n;i++)
{
ll t=a[i]*(i-1)-sum;
sum+=a[i];
ans+= t - mp[a[i]-1]+mp[a[i]+1];
mp[a[i]]++;

}
cout<<fixed<<setprecision(0)<<ans<<endl;
return 0;
}

13