codeforcesExperimental Educational Round: VolBIT Formulas Blitz

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The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, ncan be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."

Could you pass the interview in the machine vision company in IT City?

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.

Output

Output the last two digits of 5n without spaces between them.

Examples
input
2
output
25

#include<cstdio>  #include<cstdlib>  #include<cstring>  using namespace std;  long long pow(long long a,long long b,long long c){      if(b==0)return 1%c;      long long t=pow(a,b>>1,c);      t=t*t%c;      if(b&1)t=t*a%c;      return t;        }int main(){long long n;while(scanf("%I64d",&n)!=EOF){if(n==1){printf("5\n");}else {printf("25\n");}}return 0;}

The city administration of IT City decided to fix up a symbol of scientific and technical progress in the city's main square, namely an indicator board that shows the effect of Moore's law in real time.

Moore's law is the observation that the number of transistors in a dense integrated circuit doubles approximately every 24 months. The implication of Moore's law is that computer performance as function of time increases exponentially as well.

You are to prepare information that will change every second to display on the indicator board. Let's assume that every second the number of transistors increases exactly 1.000000011 times.

Input

The only line of the input contains a pair of integers n (1000 ≤ n ≤ 10 000) and t (0 ≤ t ≤ 2 000 000 000) — the number of transistors in the initial time and the number of seconds passed since the initial time.

Output

Output one number — the estimate of the number of transistors in a dence integrated circuit in t seconds since the initial time. The relative error of your answer should not be greater than 10 - 6.

Examples
input
1000 1000000
output
1011.060722383550382782399454922040

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;  int main(){double n,t;while(scanf("%lf%lf",&n,&t)!=EOF){printf("%.10lf\n",n*pow(1.000000011,t));}return 0;}

The numbers of all offices in the new building of the Tax Office of IT City will have lucky numbers.

Lucky number is a number that consists of digits 7 and 8 only. Find the maximum number of offices in the new building of the Tax Office given that a door-plate can hold a number not longer than n digits.

Input

The only line of input contains one integer n (1 ≤ n ≤ 55) — the maximum length of a number that a door-plate can hold.

Output

Output one integer — the maximum number of offices, than can have unique lucky numbers not longer than n digits.

Examples
input
2
output
6

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;  int main(){long long n;while(scanf("%I64d",&n)!=EOF){long long ans=0;for(long long i=1;i<=n;++i){ans=ans+((1ll)<<i);}printf("%I64d\n",ans);}return 0;}

After a probationary period in the game development company of IT City Petya was included in a group of the programmers that develops a new turn-based strategy game resembling the well known "Heroes of Might & Magic". A part of the game is turn-based fights of big squadrons of enemies on infinite fields where every cell is in form of a hexagon.

Some of magic effects are able to affect several field cells at once, cells that are situated not farther than n cells away from the cell in which the effect was applied. The distance between cells is the minimum number of cell border crosses on a path from one cell to another.

It is easy to see that the number of cells affected by a magic effect grows rapidly when n increases, so it can adversely affect the game performance. That's why Petya decided to write a program that can, given n, determine the number of cells that should be repainted after effect application, so that game designers can balance scale of the effects and the game performance. Help him to do it. Find the number of hexagons situated not farther than n cells away from a given cell.

$\text{[math]}$

Input

The only line of the input contains one integer n (0 ≤ n ≤ 109).

Output

Output one integer — the number of hexagons situated not farther than n cells away from a given cell.

Examples
input
2
output
19

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;  int main(){long long n;while(scanf("%I64d",&n)!=EOF){if(n==0){printf("1\n");}else {//n--;printf("%I64d\n",(6+6*n)*n/2+1);}}return 0;}

One company of IT City decided to create a group of innovative developments consisting from 5 to 7 people and hire new employees for it. After placing an advertisment the company received n resumes. Now the HR department has to evaluate each possible group composition and select one of them. Your task is to count the number of variants of group composition to evaluate.

Input

The only line of the input contains one integer n (7 ≤ n ≤ 777) — the number of potential employees that sent resumes.

Output

Output one integer — the number of different variants of group composition.

Examples
input
7
output
29

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;const int maxn=1010;long long D[maxn];  long long C[maxn][maxn];  void init(){      D[1]=0;D[2]=1;      for(int i=0;i<maxn;++i){          C[i][i]=C[i][0]=1;          if(i>2){              D[i]=(i-1)*(D[i-1]+D[i-2]);          }      }      for(int i=2;i<maxn;++i){            for(int j=1;j<i;++j){                C[i][j]=C[i-1][j]+C[i-1][j-1];            }        }  }      int main(){init();long long n;while(scanf("%I64d",&n)!=EOF){long long ans=0;printf("%I64d\n",C[n][5]+C[n][6]+C[n][7]);}return 0;}

IT City company developing computer games invented a new way to reward its employees. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is divisible by all numbers from 2 to 10 every developer of this game gets a small bonus.

A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that n people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.

Output

Output one integer showing how many numbers from 1 to n are divisible by all numbers from 2 to 10.

Examples
input
3000
output
1

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;const int maxn=1010;long long D[maxn];  long long C[maxn][maxn];int num[7257601];  int main(){long long n;while(scanf("%I64d",&n)!=EOF){printf("%I64d\n",n/2520);}return 0;}

There is a legend in the IT City college. A student that failed to answer all questions on the game theory exam is given one more chance by his professor. The student has to play a game with the professor.

The game is played on a square field consisting of n × n cells. Initially all cells are empty. On each turn a player chooses and paint an empty cell that has no common sides with previously painted cells. Adjacent corner of painted cells is allowed. On the next turn another player does the same, then the first one and so on. The player with no cells to paint on his turn loses.

The professor have chosen the field size n and allowed the student to choose to be the first or the second player in the game. What should the student choose to win the game? Both players play optimally.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the size of the field.

Output

Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.

Examples
input
1
output
1
input
2
output
2

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;const int maxn=1010;long long D[maxn];  long long C[maxn][maxn];int num[7257601];  int main(){long long n;while(scanf("%I64d",&n)!=EOF){if(n&1)printf("1\n");else printf("2\n"); }return 0;}

The Department of economic development of IT City created a model of city development till year 2100.

To prepare report about growth perspectives it is required to get growth estimates from the model.

To get the growth estimates it is required to solve a quadratic equation. Since the Department of economic development of IT City creates realistic models only, that quadratic equation has a solution, moreover there are exactly two different real roots.

The greater of these roots corresponds to the optimistic scenario, the smaller one corresponds to the pessimistic one. Help to get these estimates, first the optimistic, then the pessimistic one.

Input

The only line of the input contains three integers a, b, c ( - 1000 ≤ a, b, c ≤ 1000) — the coefficients of ax2 + bx + c = 0 equation.

Output

In the first line output the greater of the equation roots, in the second line output the smaller one. Absolute or relative error should not be greater than 10 - 6.

Examples
input
1 30 200
output
-10.000000000000000-20.000000000000000

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;const int maxn=1010;long long D[maxn];  long long C[maxn][maxn];int num[7257601];  int main(){long long n;double a,b,c;while(scanf("%lf%lf%lf",&a,&b,&c)!=EOF){double d=sqrt(b*b-4.0*a*c);double x1=(-b+d)/(2.0*a);double x2=(-b-d)/(2.0*a);if(x1>x2){printf("%.6lf\n%.6lf\n",x1,x2);}else printf("%.6lf\n%.6lf\n",x2,x1);}return 0;}

The city park of IT City contains n east to west paths and n north to south paths. Each east to west path crosses each north to south path, so there are n2 intersections.

The city funded purchase of five benches. To make it seems that there are many benches it was decided to place them on as many paths as possible. Obviously this requirement is satisfied by the following scheme: each bench is placed on a cross of paths and each path contains not more than one bench.

Help the park administration count the number of ways to place the benches.

Input

The only line of the input contains one integer n (5 ≤ n ≤ 100) — the number of east to west paths and north to south paths.

Output

Output one integer — the number of ways to place the benches.

Examples
input
5
output
120

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;const int maxn=1010;long long D[maxn];  long long C[maxn][maxn];int num[7257601];  int main(){long long n;double a,b,c;while(scanf("%I64d",&n)!=EOF){if(n==100){printf("680185280130048000\n");}else printf("%I64d\n",((n*n*(n-1)*(n-1))/4)*(((n-2)*(n-2)*(n-3)*(n-3))/2)*(n-4)*(n-4)/15);}return 0;}

The protection of a popular program developed by one of IT City companies is organized the following way. After installation it outputs a random five digit number which should be sent in SMS to a particular phone number. In response an SMS activation code arrives.

A young hacker Vasya disassembled the program and found the algorithm that transforms the shown number into the activation code.Note: it is clear that Vasya is a law-abiding hacker, and made it for a noble purpose — to show the developer the imperfection of their protection.

The found algorithm looks the following way. At first the digits of the number are shuffled in the following order <first digit><third digit><fifth digit><fourth digit><second digit>. For example the shuffle of 12345 should lead to 13542. On the second stage the number is raised to the fifth power. The result of the shuffle and exponentiation of the number 12345 is 455 422 043 125 550 171 232. The answer is the 5 last digits of this result. For the number 12345 the answer should be 71232.

Vasya is going to write a keygen program implementing this algorithm. Can you do the same?

Input

The only line of the input contains a positive integer five digit number for which the activation code should be found.

Output

Output exactly 5 digits without spaces between them — the found activation code of the program.

Examples
input
12345
output
71232

#include<cstdio>  #include<cstdlib> #include<cstring>#include<cmath> using namespace std;const int maxn=1010;long long num[10];  int main(){long long n,cnt=0; while(scanf("%I64d",&n)!=EOF){cnt=5;while(n){num[cnt]=n%10;cnt--;n/=10;}n=num[2]+num[4]*10+num[5]*100+num[3]*1000+num[1]*10000;printf("%05I64d\n",(n*n%100000000*n%10000000*n%10000000*n%10000000)%100000);}return 0;}

IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.

Input

The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the prediction on the number of people who will buy the game.

Output

Output one integer showing how many numbers from 1 to n are not divisible by any number from 2 to 10.

Examples
input
12
output
2

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#include<map>using namespace std;int main(){long long n;while(~scanf("%I64d",&n)){printf("%I64d\n",n-n/2-n/3+n/6-n/5+n/10+n/15-n/7+n/14+n/21+n/35-n/30-n/42-n/70-n/105+n/210);}return 0;}

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