Educational Codeforces Round 87 (Rated for Div. 2)日常训练

A. Alarm Clock

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.

Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.

Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he’s slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.

If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.

You just want to find out when will Polycarp get out of his bed or report that it will never happen.

Please check out the notes for some explanations of the example.

Input
The first line contains one integer t (1≤t≤1000) — the number of testcases.

The only line of each testcase contains four integers a,b,c,d (1≤a,b,c,d≤109) — the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.

Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.

Example
inputCopy
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
outputCopy
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4⋅6=27.

The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.

In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.

In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute

#include
using namespace std;
long long t,m,n,i,j,k,a,b,c,d;
int main(){
	cin>>t;
	while(t--){
		cin>>a>>b>>c>>d;
		if((b<a)&&(c<=d)){
			cout<<-1<<endl;
		}
		else if(a<=b){
			cout<<b<<endl;
		}
		else{
			m=b+((a-b)/(c-d)*c);
			if ((a-b)%(c-d)!=0) m=m+c;
			cout<<m<<endl;
		}
	}
	return 0;
} 

B. Ternary String

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given a string s such that each its character is either 1, 2, or 3. You have to choose the shortest contiguous substring of s such that it contains each of these three characters at least once.

A contiguous substring of string s is a string that can be obtained from s by removing some (possibly zero) characters from the beginning of s and some (possibly zero) characters from the end of s.

Input
The first line contains one integer t (1≤t≤20000) — the number of test cases.

Each test case consists of one line containing the string s (1≤|s|≤200000). It is guaranteed that each character of s is either 1, 2, or 3.

The sum of lengths of all strings in all test cases does not exceed 200000.

Output
For each test case, print one integer — the length of the shortest contiguous substring of s containing all three types of characters at least once. If there is no such substring, print 0 instead.

Example
inputCopy
7
123
12222133333332
112233
332211
12121212
333333
31121
outputCopy
3
3
4
4
0
0
4
Note
Consider the example test:

In the first test case, the substring 123 can be used.

In the second test case, the substring 213 can be used.

In the third test case, the substring 1223 can be used.

In the fourth test case, the substring 3221 can be used.

In the fifth test case, there is no character 3 in s.

In the sixth test case, there is no character 1 in s.

In the seventh test case, the substring 3112 can be used.

#include
#include
#include
using namespace std;
string s;
int t,i,j,k,m,n,l,ans;
int p[10];
int main(){
	cin>>t;
	while(t--){
		cin>>s;
		l=s.length();
		ans=l;
		p[0]=p[1]=p[2]=-1;
		for(i=0;i<l;i++){
			p[s[i]-'1']=i;
			if(p[0]!=-1 && p[1]!=-1 && p[2]!=-1){
				ans=min(ans,max(max(p[0],p[1]),p[2])-min(min(p[0],p[1]),p[2])+1);
			}
		}
		if(p[0]==-1 || p[1]==-1 || p[2]==-1) ans=0;
		cout<<ans<<endl;		
	}
	return 0;
}

C1. Simple Polygon Embedding

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
The statement of this problem is the same as the statement of problem C2. The only difference is that, in problem C1, n is always even, and in C2, n is always odd.

You are given a regular polygon with 2⋅n vertices (it’s convex and has equal sides and equal angles) and all its sides have length 1. Let’s name it as 2n-gon.

Your task is to find the square of the minimum size such that you can embed 2n-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n-gon should also lie inside or on a border of the square.

You can rotate 2n-gon and/or the square.

Input
The first line contains a single integer T (1≤T≤200) — the number of test cases.

Next T lines contain descriptions of test cases — one per line. Each line contains single even integer n (2≤n≤200). Don’t forget you need to embed 2n-gon, not an n-gon.

Output
Print T real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn’t exceed 10−6.

Example
inputCopy
3
2
4
200
outputCopy
1.000000000
2.414213562
127.321336469

#include
using namespace std;
int t,n;
int main(){
	cin>>t;
	while(t--){
		cin>>n;
		printf("%.7f\n",1/tan(asin(1)/n));
	}
	return 0;
}

后记

第一次认真做educational的题目，总体感觉数学思维还是需要的。
今天特地看了一下提交列表，发现Python已经占绝大多数了。。。看来C++也快成过时的了。的确Python功能更强大一些，代码量可以压缩的更少，更加智慧一点。