AtCoder Beginner Contest 258（A-D）

A - When?

The main points of ： Format output time

#include<bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int N = 1e5+7; 
int n;
int main()
{
    
	cin>>n;
	int h=n/60+21;
	int m=n%60;
	printf("%02d:%02d",h,m);
}

B Number Box

Be careful ：

Note that the search is in eight straight directions , Instead of searching eight directions every time , Because no matter how to search the string length, it is certain , So we only start with the largest number at the beginning

Pay attention to the questions

#include<bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int N = 1e5+7; 
 
int n;
 
char a[21][21];
int max1;
string ans="";
 
int dir[8][2]={
    0,1,0,-1,1,0,-1,0,1,1,1,-1,-1,1,-1,-1};
 
void serch(int x,int y)
{
    
	string s1="";
	int xx=x;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[xx][y];
		xx++;
		if(xx>n) xx=1;
	}// Right 
	ans=max(ans,s1);
	
	s1="";
	xx=x;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[xx][y];
		xx--;
		if(xx<1) xx=n;
	}// Left 
	ans=max(ans,s1);
	
	s1="";
	int yy=y;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[x][yy];
		yy++;
		if(yy>n) yy=1;
	}// On 
	ans=max(ans,s1);
	
	s1="";yy=y;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[x][yy];
		yy--;
		if(yy<1) yy=n;
	}// Next 
	ans=max(ans,s1);
	
	s1="";xx=x,yy=y;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[xx][yy];
		xx--;
		yy--;
		if(xx<1) xx=n;
		if(yy<1) yy=n;
	}// Top left 
	ans=max(ans,s1);
	
	s1="";xx=x,yy=y;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[xx][yy];
		xx++;
		yy++;
		if(xx>n) xx=1;
		if(yy>n) yy=1;
	}// The lower right 
	ans=max(ans,s1);
	
	s1="";xx=x,yy=y;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[xx][yy];
		xx++;
		yy--;
		if(xx>n) xx=1;
		if(yy<1) yy=n;
	}// The upper right 
	ans=max(ans,s1);
	
	s1="";xx=x,yy=y;
	for(int i=1;i<=n;i++)
	{
    
		s1+=a[xx][yy];
		xx--;
		yy++;
		if(xx<1) xx=n;
		if(yy>n) yy=1;
	}// The lower left 
	ans=max(ans,s1);
}
 
int main()
{
    
	cin>>n;
	
	for(int i=1;i<=n;i++) ans+='1';
	
	for(int i=1;i<=n;i++)
	{
    
		for(int j=1;j<=n;j++)
		{
    
			cin>>a[i][j];	
			max1=max(max1,a[i][j]-'0');
		}	
	}	
	
	for(int i=1;i<=n;i++)
	{
    
		for(int j=1;j<=n;j++)
		{
    
			if(a[i][j]-'0'==max1)
			{
    
				serch(i,j);
			}
		}
	}
	
	
	cout<<ans;
	
	
}

C Rotation

Ideas ：

We use it k Represents the number of shift operations , When k == n Time is equivalent to no movement Then the number of effective moves is k%n , Then the original header is equivalent to being moved to n-k The location of , Requirement No b Letters , That's it n-k+b The letter of position , Of course , This number should also be correct n Remainder , If you exceed n
Represents that this number has been moved to the front , So take the rest .

Be sure to note that the remainder equals 0 Does the situation of have an impact on the problem ！！！！
Try more groups of examples ！！！
Pay attention long long

#include<bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const int N = 1e5+7; 
 
ll n,m;
string s;
ll a,b,k,sum;
 
int main()
{
    
	cin>>n>>m;
	cin>>s;
	for(int i=1;i<=m;i++)
	{
    
		scanf("%lld %lld",&a,&b);
		if(a==1)
		{
    
			k+=b;
			k=k%n;
		}
		else
		{
    
			int kk=((n-k)+b)%n;if(kk==0) kk=n;
			cout<<s[kk-1]<<endl;
		}
 
	}
}

D:Trophy

The question ：

n Stages , From the first stage , Each new stage will cost a+b , Staying in the unlocked phase only costs b , ask Handle X What is the minimum cost of the second stage

Ideas ：

We need to find a smaller one b, Make most of The treatment of is in b On , Not in a+b On , Make the cost smaller , So the idea of this question is to unlock each stage , Then process all the remaining processing times in this stage, that is, spend this b , At every stage b Try it all once , Minimum value .

Note that the data is relatively large , open ull

#include<bits/stdc++.h>
using namespace std;
 
typedef unsigned long long ll;
const int N = 1e5+7; 
 
ll min1=1e19+10;
 
ll n,t;
ll a,b;
ll sum;
 
int main()
{
    
	cin>>n>>t;
	
	for(int i=1;i<=n;i++)
	{
    
		scanf("%llu %llu",&a,&b);
		sum+=(a+b);
		min1=min(min1,sum+b*(t-i));
	}
	
	cout<<min1;
}