牛客小白月赛62题解

A.幼稚园的树

模拟一下即可，注意第m天中午就是第m-1的最终状态

/*
    悲观看待成功，乐观看待失败。 
    author：leimingze 
*/
#include
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
const int base=131;
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'//交互题删掉此 
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){return (a%mod+mod)%mod;}
int lowbit(int x){return x&-x;}//最低位1及其后面的0构成的数值
int qmi(int a, int k, int p){int res = 1 % p;while (k){if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
int inv(int a,int mod){return qmi(a,mod-2,mod);}
int lcm(int a,int b){return a*b/__gcd(a,b);}
int n;
const int N=1010;
int h[N];
int a,k,b,m;
void solve()
{
	cin>>n;
	rep(i,1,n)cin>>h[i];
	cin>>a>>k>>b;
	cin>>m;
	rep(i,1,m-1)
		rep(j,1,n)
		{
			h[j]+=a;
			if(h[j]>k)h[j]=b;
		}
	rep(i,1,n)cout<<h[i]<<' ';
	cout<<endl;
}
signed main() 
{
	io;
	int _;_=1;
	cin>>_;
	while(_--)solve();
	return 0;
}

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B.剩下的数

区间和为sum
令 y＝sum%x
从l到r挨个%x后 必定有一个数z%x等于y(因为x≤(r-l+1))
(sum-z) % x＝0
则最后只会剩下这个z
上面讨论的是sum%x != 0的情况
sum%x=0直接就0就好了

/*
    悲观看待成功，乐观看待失败。 
    author：leimingze 
*/
#include
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
const int base=131;
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'//交互题删掉此 
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){return (a%mod+mod)%mod;}
int lowbit(int x){return x&-x;}//最低位1及其后面的0构成的数值
int qmi(int a, int k, int p){int res = 1 % p;while (k){if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
int inv(int a,int mod){return qmi(a,mod-2,mod);}
int lcm(int a,int b){return a*b/__gcd(a,b);}
int n;
int l,r;
int m;
void solve()
{
	cin>>l>>r;
	cin>>m;
	int s=(l+r)*(r-l+1)/2;
	while(m--)
	{
		int x;
		cin>>x;
		if(s%x)cout<<1<<endl;
		else cout<<0<<endl;
	}
}
signed main() 
{
	io;
	int _;_=1;
	cin>>_;
	while(_--)solve();
	return 0;
}

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C.数组划分

对于A数组分解质因数，然后用数组B的每个元素的质因子看是否在A中能找到

/*
    悲观看待成功，乐观看待失败。 
    author：leimingze 
*/
#include
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
const int base=131;
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'//交互题删掉此 
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){return (a%mod+mod)%mod;}
int lowbit(int x){return x&-x;}//最低位1及其后面的0构成的数值
int qmi(int a, int k, int p){int res = 1 % p;while (k){if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
int inv(int a,int mod){return qmi(a,mod-2,mod);}
int lcm(int a,int b){return a*b/__gcd(a,b);}
int n;
const int N=1e5+10;
int a[N],b[N];
map<int,int>mp;
bool get(int x,bool f)
{
	if(!f)
	{
		int m=x;
		rep(i,2,x/i)
		{
			if(x%i==0)
			{
				while(x%i==0)
					x/=i;
				mp[i]++;
			}
		}
		if(x>1)mp[x]++;
	}
	else
	{
		rep(i,2,x/i)
		{
			if(x%i==0)
			{
				while(x%i==0)
					x/=i;
				if(mp.count(i))return false;
			}
		}
		if(x>1&&mp.count(x))return false;
	}
	return true;
}
void solve()
{
	cin>>n;
	rep(i,1,n)cin>>a[i];
	rep(i,1,n)cin>>b[i];
	rep(i,1,n)
		get(a[i],0);
	rep(i,1,n)
	{
		if(!get(b[i],1))
		{
			cout<<"No"<<endl;
			return;
		}
	}
	cout<<"Yes"<<endl;
}
signed main() 
{
	io;
	int _;_=1;
	//cin>>_;
	while(_--)solve();
	return 0;
}

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D.子树的大小

题目隐含条件是这棵树是一颗完全二叉树。
那么这道题就可以转换成一道数学题了。
一棵满K叉树的结点个数：(1-k^n)/(1-k)

/*
    悲观看待成功，乐观看待失败。 
    author：leimingze 
*/
#include
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
const int base=131;
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'//交互题删掉此 
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){return (a%mod+mod)%mod;}
int lowbit(int x){return x&-x;}//最低位1及其后面的0构成的数值
//int qmi(int a, int k, int p){int res = 1 % p;while (k){if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
//int inv(int a,int mod){return qmi(a,mod-2,mod);}
int lcm(int a,int b){return a*b/__gcd(a,b);}
int qmi(int a, int b){int res = 1;while(b){if(b & 1) res *= a;a *= a;b >>= 1;}return res;}
int n,k,m;
const int N=1e5+10; 
int q[N];
void solve()
{
	cin>>n>>k>>m;
	rep(i,1,m)cin>>q[i];
	rep(i,1,m)
	{
		if(k==1)
		{
			cout<<n-q[i]<<endl;
			continue;
		}
		int x=q[i];
		int hl=0;
		while(x<=n-1)
		{
			hl++;
			x=x*k+1;
		}
		x=(x-1)/k;
		int y=q[i];
		int hr=0;
		while(y<=n-1)
			y=k*y+k,hr++;
		y/=(k+1);
		int res=(qmi(k,hr)-1)/(k-1);
		if(hr<hl)res+=n-x;
		cout<<res<<endl;
	}
}
signed main() 
{
	io;
	int _;_=1;
	cin>>_;
	while(_--)solve();
	return 0;
}

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E.剖分

树形差分维护一下即可
注意前缀和时不是回溯，是递归

/*
    悲观看待成功，乐观看待失败。 
    author：leimingze 
*/
#include
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
const int base=131;
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'//交互题删掉此 
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){return (a%mod+mod)%mod;}
int lowbit(int x){return x&-x;}//最低位1及其后面的0构成的数值
int qmi(int a, int k, int p){int res = 1 % p;while (k){if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
int inv(int a,int mod){return qmi(a,mod-2,mod);}
int lcm(int a,int b){return a*b/__gcd(a,b);}
int n,m;
const int N=1e7+10;
int a[N],b[N];
int s[N]; 
int cnt[N];
int dfs(int u,int sum)
{
	sum+=a[u];
	if(u*2<=n)s[u]+=dfs(u*2,sum);
	if(u*2+1<=n)s[u]+=dfs(u*2+1,sum);
	s[u]=b[u]+sum;
	return s[u];
}
void solve()
{
	cin>>n>>m;
	rep(i,1,m)
	{
		int op,x;
		cin>>op>>x;
		if(op==1)a[x]++;
		else if(op==2)a[1]++,a[x]--;
		else if(op==3)
			while(x>=1)b[x]++,x/=2;
		else
		{
			a[1]++;
			while(x>=1)b[x]--,x/=2;
		}
	}
	dfs(1,0);
	rep(i,1,n)cnt[s[i]]++;
	rep(i,0,m)cout<<cnt[i]<<' ';
	cout<<endl;
}
signed main()
{
	io;
	int _;_=1;
	//cin>>_;
	while(_--)solve();
}
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