【A-D】Educational Codeforces Round 133 (Rated for Div. 2)参考代码

本文提供了Educational Codeforces Round 133 (Rated for Div. 2)比赛中的A-D题目的解题思路和C++参考代码,包括2-3次移动问题、排列链、走廊中的机器人以及芯片移动问题。

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A. 2-3 Moves

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#define ll long long
using namespace std;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}

void solve()
{
	ll n=read();
	if(n==1)
	{
		cout<<2<<endl;
		return;
	}
	if(n==3)
	{
		cout<<1<<endl;
		return;
	}
	ll ans=0;
	if(n>4)ans+=n/3,n%=3;
	if(n%2!=0)ans--,n+=3;
	ans+=n/2;
	printf("%lld\n",ans);
}

int main()
{
	int T=read();
	while(T--){solve();}
	return 0;
}

B. Permutation Chain

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#define ll long long
using namespace std;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}
const int N=110;
int a[N];
void solve()
{
	int n=read();
	for(int i=1;i<=n;i++)a[i]=i;
	printf("%d\n",n);
	for(int i=1;i<=n;i++)printf("%d ",a[i]);
	printf("\n");
	for(int i=1;i<=n-1;i++)
	{
		swap(a[i],a[i+1]);
		for(int j=1;j<=n;j++)printf("%d ",a[j]);
		printf("\n");
	}
}

int main()
{
	int T=read();
	while(T--){solve();}
	return 0;
}

C. Robot in a Hallway

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<set>
#include<string>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#define ll long long
using namespace std;
const int mod=1e9+7;
const int INF=0x3f3f3f3f;

inline int read()
{
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0',c=getchar();}
    return x*f;
}
const int N=2e5+10;
   ll a[2][N];
    ll f[2][N];
void solve()
{
    memset(f,0,sizeof(f));
	int n=read();
	for(int i=0;i<2;i++)
		for(int j=0;j<=n-1;j++)scanf("%lld",&a[i][j]);
	a[0][0]=-1;
	
	for(int i=n-1;i>=0;i--)
	{
		for(int j=0;j<=1;j++)
		{
			f[j][i]=max(a[j][i]+1,f[j][i+1]-1);
			int p;
			if(j==0)p=1;
			else p=0;
			f[j][i]=max(f[j][i],a[p][i]-2*n+2*i+2);
		}
	}
	ll ans=1e18,fuck=0;
	int x=0,y=0; 
	for(int i=0;i<n;i++)
	{
		ll mx=max(fuck,f[x][y])+2*(n-i)-1;
		ans=min(ans,mx);
		if(x==0)x=1;
		else x=0;
		fuck=max(a[x][y]+1,fuck+1);
		y++;
		if(y==n)ans=min(ans,fuck);
		else fuck=max(a[x][y]+1,fuck+1);
	}
	printf("%lld\n",ans);
}

int main()
{
	int T=read();
	while(T--){solve();}
	return 0;
}

D. Chip Move

void solve(){
	cin >> n >> k;
	f[k][0] = 1;

	for(int j = 0, ff = 0; j * j <= 2 * n; j ++, ff ^= 1)
	{
		for(int i = 0; i <= n; i ++)
			f[i][ff ^ 1] = 0;
		for(int i = 1; i <= n; i ++)
			if(i - j - k > 0)
			{
				(f[i][ff] += f[i - j -k][ff]) %= Mod;
				if(i - j - k > 0)(f[i][ff ^ 1]+=f[i - j - k - 1][ff]) %= Mod;
			}
		for(int i = 1; i <= n; i ++){
			(ans[i] += f[i][ff]) %= Mod;
		}
	}
	for(int i = 1; i <= n; i ++){
		cout << ans[i] % Mod << " ";
	}
}
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