蓝桥算法训练营—提高组—2023年2.12日

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P1890 gcd区间

P1890 gcd区间 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include <bits/stdc++.h>
#define x first
#define y second
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef pair<int,int> PII;
typedef pair<char,int> PCI;
typedef long long LL;
typedef unsigned long long ULL;

const int N=1000+10 ,M= N * N + 10, INF = 0x3f3f3f3f ,mod =1e9 + 7 ;


int n,m;


struct node{
	int l,r,v;
}tr[N * 4 ];


int gcd(int a,int b)
{
	return b?gcd(b,a%b):a;
}

void pushup(int u)
{
	tr[u].v = gcd(tr[u<<1].v,tr[u<<1|1].v);
}
void build(int u,int l,int r)
{
	if(l == r) {
		int t;cin>>t;
		tr[u]={l,r,t};
	}
	else 
	{
		tr[u] = {l,r};
		int mid = l + r >> 1;
		build(u<<1,l,mid);
		build(u<<1|1,mid+1,r);
		pushup(u);
	}
	
}

int query(int u,int l,int r)
{
	if(tr[u].l >= l && tr[u].r <= r)
	{
		return tr[u].v;
	}
	
	int mid = tr[u].l + tr[u].r >> 1 ; 
	
	int v=0;
	if(l<=mid ) v =query(u<<1,l,r);
	if(r > mid ) v =gcd(v,query(u<<1|1,l,r));
	
	return v;
}



void solve()
{
	cin>>n>>m;
	build(1,1,n);	
	
	while(m -- )	
	{
		int x,y;cin>>x>>y;
		cout << query(1,x,y) << endl;
	}
}


int main()
{
	ios 
	int T=1;
// 	cin>>T;
	while(T -- )
	{
		solve();
	}	
	
	
	
	
	return 0;
}

P1405 苦恼的小明

P1405 苦恼的小明 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

#include <bits/stdc++.h>
#define x first
#define y second
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef pair<int,int> PII;
typedef pair<char,int> PCI;
typedef long long LL;
typedef unsigned long long ULL;

const int N=1e7+10 , INF = 0x3f3f3f3f ;


int n,m;

int phi(int x)
{
	LL res =x ;
	for(int i=2;i<=x/i;i++)
	if(x % i ==0 ) 
	{
		res = res /i * (i-1);
		while(x % i == 0) x /=i;
	}
	if(x > 1) res = res / x * (x-1) ;
	return res;
}
int gcd(int a,int b)
{
	return b?gcd(b,a%b):a;
}
LL qmi(LL a,LL b,int mod)
{
	LL res =1;
	while(b)
	{
		if(b & 1) res = (res * a) % mod;
		a = (a * a ) % mod;
		b>>=1;
	}
	return res;
}

int a[N];

int dfs(int x,int mod)
{
	if(x == n ) return a[x]=a[x] % mod; 
	
	LL t = phi(mod);
	LL t1=dfs(x+1,t);
	
	if(gcd(a[x],mod) == 1 )  
	return a[x] = qmi(a[x],t1,mod);
	else 
	{
		if(a[x+1] > t ) return a[x] =qmi(a[x],t1+t,mod);
		return a[x] = qmi(a[x],a[x+1],mod);
	}
	
	
}



void solve()
{
	cin>>n;
	for(int i=1;i<=n;i ++ ) cin>> a[i];
	int mod=10007;
	cout << dfs(1,mod) << endl;
}

int main()
{
	ios 
	LL T=1;
// 	cin>>T;
	while(T -- )
	{
		solve();
	}	
	
	
	
	
	return 0;
}