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, n can 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?
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 the last two digits of 5n without spaces between them.
2
25
题解：我用快速幂写了下，其实感觉都是25的；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; int main(){ LL n; while(~scanf("%I64d",&n)){ int ans=1,k=5; while(n){ if(n&1)ans=k*ans%100; else k=k*k%100; n>>=1; } printf("%d ",ans); } 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.
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 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.
1000 1000000
1011.060722383550382782399454922040
题解：
n*pow(1.000000011,t)，我竟然还想着快速幂；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; int main(){ LL n,t; while(~scanf("%I64d%I64d",&n,&t)){ printf("%.15f ",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.
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 one integer — the maximum number of offices, than can have unique lucky numbers not longer than n digits.
2
6
题解：很好找的规律，按照位置来看，每个位置有两种选择7或者8；直接从1到n把所有情况加起来；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; int main(){ int n; while(~SI(n)){ LL ans=0; LL cur=1; for(int i=1;i<=n;i++){ cur*=2; ans+=cur; } printf("%I64d ",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.
The only line of the input contains one integer n (0 ≤ n ≤ 109).
Output one integer — the number of hexagons situated not farther than n cells away from a given cell.
2
19
题解：题读了半天，规律也找了好一会，然后超时，最后LL又挂了两次。。
for(int i=1;i<=n;i++){
cur++;
last++;
ans+=(cur*2+last*4+2);
}
规律总结下就是：(LL)(n+3)*n+(LL)n*(n-1)*2+(LL)2*n+1;
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; int main(){ int n; while(~SI(n)){ //LL cur=1,last=-1,ans=1; LL ans=(LL)(n+3)*n+(LL)n*(n-1)*2+(LL)2*n+1; /*for(int i=1;i<=n;i++){ cur++; last++; ans+=(cur*2+last*4+2); }*/ printf("%I64d ",ans); } 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.
The only line of the input contains one integer n (7 ≤ n ≤ 777) — the number of potential employees that sent resumes.
Output one integer — the number of different variants of group composition.
7
29
题解：规律很好推，关键在于LL上，应该每次乘直接除，要不会超LL
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; LL select(LL n){ LL ans=0; for(int i=5;i<=7;i++){ LL temp=1; for(int j=1;j<=i;j++){ temp=temp*(n-j+1)/j; } ans+=temp; } return ans; } int main(){ LL n; while(~scanf("%I64d",&n)){ LL ans=select(n); printf("%I64d ",ans); } 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.
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 one integer showing how many numbers from 1 to n are divisible by all numbers from 2 to 10.
3000
1
题解：直接找1到10的最小公倍数除以n取整就可以了；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; LL gcd(LL a,LL b){ return b==0?a:gcd(b,a%b); } int main(){ LL n; while(~scanf("%I64d",&n)){ LL lcm=1; for(LL i=1;i<=10;i++){ lcm=lcm/gcd(lcm,i)*i; } printf("%I64d ",n/lcm); } 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.
The only line of the input contains one integer n (1 ≤ n ≤ 1018) — the size of the field.
Output number 1, if the player making the first turn wins when both players play optimally, otherwise print number 2.
1
1
2
2
题解：
谁先赢，奇偶喽
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; int main(){ LL n; while(~scanf("%I64d",&n)){ printf("%d ",n&1?1:2); } 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.
The only line of the input contains three integers a, b, c ( - 1000 ≤ a, b, c ≤ 1000) — the coefficients of ax2 + bx + c = 0equation.
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.
1 30 200
-10.000000000000000 -20.000000000000000
由于a正负不确定需要max，min；
代码：
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> using namespace std; const int INF=0x3f3f3f3f; #define mem(x,y) memset(x,y,sizeof(x)) #define SI(x) scanf("%d",&x) #define PI(x) printf("%d",x) #define SD(x) scanf("%lf",&x) #define P_ printf(" ") typedef __int64 LL; int main(){ double a,b,c; while(~scanf("%lf%lf%lf",&a,&b,&c)){ double t=sqrt(b*b-4*a*c); printf("%.15f %.15f ",max((-b+t)/(2*a),(-b-t)/(2*a)),min((-b+t)/(2*a),(-b-t)/(2*a))); } return 0; }