2020.10.30【CSP-J/S】Summary of popular group simulation competition

Question number	topic
T1	Straight line on the plane
T2	Interesting water pipe
T3	Homework Questions for Little C
T4	Rent a car for a spring outing
T5	Row numbers
Score	185/205

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T1

Ideas

Not made.

T2

Ideas

This question is a dichotomy, it can be done $O({\log }_{2}k)$ .
We can start from$k=In$ the case of$1$ , start pushing the formula.
From$1☞k-1$ two points a$mid$ ,mid midcan be obtained from thesum formula of arithmetic sequence$mid$ 的贡献是：
$$(mid\ast k)-(mid-1)-\frac{(mid-1)\ast mid}{2}$$
For $>=Just$ maintain the answer for$n$ .

Code:

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
long long n,k,js,ans=-1; 
int main()
{
    
    
	cin>>n>>k;
	if(n==1)
	 {
    
    
	 	cout<<0;
	 	return 0;
	 }
	if(n<=k)
	 {
    
    
	 	cout<<1;
	 	return 0;
	 }
	long long l=1,r=k-1,mid=0;
	while(l<=r)
	 {
    
    
	 	mid=(l+r)/2;
	 	js=(mid*k)-(mid-1)-(mid-1)*mid/2;
	 	if(js>=n)
	 	  r=mid-1,ans=mid;
	 	else
	 	  l=mid+1;
	 }
	cout<<ans;
	return 0;
}

T3

Ideas

Not made.

T4

Ideas

Not made

T5

Ideas

For game theory , take the second largest of each row.
For specific ideas, please see the first problem solution of luogu P1199 Three Kingdoms Game

Code

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,a[1010][1010],ans;
int main()
{
    
    
	cin>>n;
	for(int i=1; i<n; i++)
	 for(int j=i+1; j<=n; j++)
	  	cin>>a[i][j],a[j][i]=a[i][j];
	for(int i=1; i<=n; i++)
	  {
    
    
	    sort(a[i]+1,a[i]+1+n);
	    ans=max(ans,a[i][n-1]);
	  }
	cout<<1<<endl<<ans;
	return 0;
}