[Ybt] [Basic calculation two-point class example 2] Armor arrangement

Armor layout

Topic link: Arrangement of armor

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Title description

Problem solving ideas

First of all, how many armors can be found for each point and the previous point, $O\left(n\right)$。

Then, because only one line of defense has a problem, if the current point is an odd number, then the line of defense in question must be before it, and we can divide it based on this. $O(\log n)$。

Of course, if the $214748364\; The7$ lines of defense are even, so all the lines of defense are fine.

code

#include<iostream>
#include<cstdio>
#define int long long
using namespace std;

int T;
int n;
int s[200010],e[200010],d[200010];

int check(int t)
{
    
    
	int ans=0;
	for(int i=1;i<=n;i++)
		if(s[i]<=t)
			ans+=(min(e[i],t)-s[i])/d[i]+1;
	return ans;
}

void work()
{
    
    
	if(check(2147483647)%2==0)
	{
    
    
		printf("There's no weakness.\n");
		return; 
	}
	int l=0,r=2147483647;
	while(l<r)
	{
    
    
		int mid=(l+r)/2;
		if(check(mid)%2==1)
			r=mid;
		else
			l=mid+1;
	}
	cout<<l<<" "<<check(l)-check(l-1)<<endl;
}

signed main()
{
    
    
	cin>>T;
	while(T--)
	{
    
    
		cin>>n;
		for(int i=1;i<=n;i++)
			scanf("%lld%lld%lld",&s[i],&e[i],&d[i]);
		work();
	}
}