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两只塔姆沃斯牛 The Tamworth Twohttps://www.luogu.com.cn/problem/P1518

思路：大模拟

#include<stdio.h> 
#include <stdlib.h>
#include<bits/stdc++.h>
#define itn int
using namespace std;
int main()
{
	int mx[4]={-1,0,1,0};
	int my[4]={0,1,0,-1};
	int d1=0,d2=0;
	char a[11][11];
	int x1,x2,y1,y2;
	for(int i=0;i<10;++i)
	{
		scanf("%s",&a[i]);
	}
	for (int i=0;i<10;++i)
	{
		for (int j=0;j<10;++j)
		{
			if (a[i][j]=='F')//F是人 
		{
			x1=i,y1=j;
		}
		else if (a[i][j]=='C')
		{
			x2=i,y2=j;
		}	
		}
	}
	bool visited[10][10][4][10][10][4] = {};
	int time =0;
	while (x1!=x2 || y1!=y2)
	{
		if (visited[x1][y1][d1][x2][y2][d2]) {
        cout << "0";
        return 0;
    	}
		visited[x1][y1][d1][x2][y2][d2] = true;	
		int tx1=x1+mx[d1],ty1=y1+my[d1];
		int tx2=x2+mx[d2],ty2=y2+my[d2];
		if (tx1>=0 && tx1<10 && ty1>=0 && ty1<10 && a[tx1][ty1]!='*')
		{	
			x1=tx1,y1=ty1;		
		}
		else 
		{
			d1=(d1+1)%4;
		}
		if (tx2>=0 && tx2<10 && ty2>=0 && ty2<10 && a[tx2][ty2]!='*')
		{			
			x2=tx2,y2=ty2;
		}
		else 
		{
			d2=(d2+1)%4;
		}
		time++;
	}
	cout<<time;
}

EKO / 砍树https://www.luogu.com.cn/problem/P1873

思路：简单二分，找到能满足题意的，最小长度

#include<bits/stdc++.h>
#define itn int
using namespace std;
long long n,m;
long long a[1000008];
int main()
{
	cin>>n>>m;
	for (int i=0;i<n;++i)cin>>a[i];
	sort(a,a+n);
	long long l=0,r=a[n-1];
	while (l<=r)
	{
		long long sum=0;
		long long mid=l+(r-l)/2;
		for (int i=0;i<n;++i)
		{
			if (a[i]>mid) sum=sum+(a[i]-mid);
		}
		if (sum>=m)l=mid+1;
		else if (sum<m) r=mid-1;
	}
	cout<<l-1;
	return 0;
}

木材加工https://www.luogu.com.cn/problem/P2440

思路：二分

#include<stdio.h> 
#include <stdlib.h>
#include<bits/stdc++.h>
#define itn int
using namespace std;
int n;
long long l=0,r=100000001,a[100005],k;
bool blbl(long long  mid)
{
	int ans=0;
	for (int i=0;i<n;++i)
	{
		ans+=a[i]/mid;
	}
	if (ans>=k)return true;
	else return false;
}
int main()
{
	cin>>n>>k;
	for (int i=0;i<n;++i)cin>>a[i];
	sort(a,a+n);
	while (l<r)
	{
		long long mid=l+(r-l)/2;
		if (blbl(mid))l=mid+1;
		else r=mid-1;	
	}
	if (l==0)
	{
		cout<<0;
		return 0;
	}
	cout<<l-1;
	
}