质因数相关的几道好题

1

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int maxn = 4e7 + 10;
bool vis[maxn];ll primes[maxn];
ll sum[maxn];
ll cnt;
int  main()
{
	
	for(ll i=2;i<maxn;i++)
	{
		if(!vis[i])
		{
			primes[++cnt]=i;
		}
		for(ll j=1;primes[j]<=maxn/i;j++)
		{
			vis[i*primes[j]]=1;
			if(i%primes[j]==0) break;
		}
	}
	for(ll i=1;i<=cnt;i++)
	{
		sum[i]=sum[i-1]+primes[i];
	}
	ll tt;cin>>tt;
	while(tt--)
	{
		ll n;cin>>n;ll ans=0;
		for(ll  i=0;i<=cnt&&primes[i]<=n;i++)
		{
			ll x = n + sum[i];
			ll pos = lower_bound(sum+1,sum+1+cnt,x)-sum;
			if(sum[pos]==x) ans++;
		}
		cout<<ans<<"\n";
	}
}

2

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1000010;

int primes[N], cnt;
bool st[N];

void init(int n)
{
    memset(st, 0, sizeof st);
    cnt = 0;
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[i * primes[j]] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    int l, r;
    int n;
    cin>>n;
    while (n--)
    {
        init(50000);
        cin>>l>>r;
        memset(st, 0, sizeof st);
        for (int i = 0; i < cnt; i ++ )
        {
            LL p = primes[i];
            for (LL j = max(p * 2, (l + p - 1) / p * p); j <= r; j += p)
                st[j - l] = true;
        }

        cnt = 0;
        for (int i = 0; i <= r - l; i ++ )
            if (!st[i] && i + l >= 2)
                primes[cnt ++ ] = i + l;

        if (cnt < 2) puts("There are no adjacent primes.");
        else
        {
            int minp = 0, maxp = 0;
            for (int i = 0; i + 1 < cnt; i ++ )
            {
                int d = primes[i + 1] - primes[i];
                if (d < primes[minp + 1] - primes[minp]) minp = i;
                if (d > primes[maxp + 1] - primes[maxp]) maxp = i;
            }

            printf("%d,%d are closest, %d,%d are most distant.\n",
                primes[minp], primes[minp + 1],
                primes[maxp], primes[maxp + 1]);
        }
    }

    return 0;
}

3

#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 3e4+5;
int p[N],idx;
bool st[N];
int qsm(int a,int b){
    int s=1;
    while(b){
        if(b&1)s=s*a;
        a=a*a;
        b>>=1;
    }
    return s;
}
void init(int n){
    for(int i=2;i<=n;i++){
        if(!st[i])p[idx++]=i;
        for(int j=0;p[j]<=n/i;j++){
            st[p[j]*i]=true;
            if(i%p[j]==0)break;
        }
    }
}
int main(){
    init(N-1);
    int T;
    scanf("%d",&T);
    while(T--){
        int k;
        scanf("%d",&k);
        int n=1;
        while(k--){
            int a,b;
            scanf("%d%d",&a,&b);
            n*=qsm(a,b);
        }
       // cout<<n<<"** "<< endl;
        n--;
        vector<int>va,vb;
        for(int i=2;i<=n/i;i++){
            if(n%i==0)
            {            
            int s = 0;
            while (n % i == 0) n /= i, s ++ ;
            if(s){
                 va.push_back(i);
                 vb.push_back(s);
                 }
            }
        }
        if(n>1)va.push_back(n);
        vb.push_back(1);
        for(int i=va.size()-1;i>=0;i--){
            printf("%d %d",va[i],vb[i]);
            if(!i)printf("\n");
            else printf(" ");
        }
    }
    return 0;
}

4

#include <bits/stdc++.h>
using namespace std;
const int N=1e6+10;
int t,x;
long long fact[N];
int main()
{
	fact[0]=1;
	for(int i=1;i<=20;i++)fact[i]=fact[i-1]*i;
	cin>>t;
	while(t--){
		unordered_map<int,int> mp;
		scanf("%d",&x);long long res=0;
		for(int i=2;i*i<=x;i++){
			while(x%i==0){
				mp[i]++;x/=i;res++;
			}
		}
		if(x>1) mp[x]++,res++;
		printf("%d ",res);
		res=fact[res];
		for(auto k:mp){
            int u=k.second;
			res/=fact[u];           
		}
		printf("%lld\n",res);
	}
}