A-青蛙的約會_牛客競賽數學專題班整數分解與篩法 (nowcoder.com)

根據題意推出表示式，用擴充套件歐幾里得可解，注意處理大小寫

#include <bits/stdc++.h>
#define fi first
#define int long long
#define se second
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
ll exgcd(ll a,ll b,ll &x,ll &y) {
	if (b==0) {
		x = 1, y = 0;
		return a;
	}
	ll c = exgcd(b,a%b,x,y);
	ll tx = x;
	x = y;
	y = tx - (a/b)*y;
	return c;
}
ll gcd(ll a,ll b) {
	if (a<b) swap(a,b);
	while (b>0) {
		ll t = a;
		a = b;
		b = t % b;
	}
	return a;
}
signed main() {
	ll x,y,m,n,L;
	cin>>x>>y>>m>>n>>L;
	ll a = L;
	ll b = n-m;
	ll c = x-y;
	if (b<0) {
		b = -b;
		c = -c;
	}
	ll d = gcd(a,b);
//	cout<<a<<" "<<b<<" "<<c<<" "<<d<<'\n';
	if (c%d!=0) cout<<"Impossible";
	else {
		exgcd(a,b,x,y);
		y = (c/d*y%(a/d)+(a/d))%(a/d);
		cout<<y;
	}
	
}

B-Sum of Consecutive Prime Numbers_牛客競賽數學專題班整數分解與篩法 (nowcoder.com)

處理出素數用尺取法

#include <bits/stdc++.h>
#define fi first
#define se second
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const int maxn = 4e7+7;
int prime[maxn];
int vis[maxn];
int tot = 0;
void getPrime(int n) {
	for (int i=2;i<=n;i++) {
		if (!vis[i]) prime[++tot] = i;
		for (int j=1;j<=tot;j++) {
			if (i*prime[j]>n) break;
			vis[i*prime[j]] = 1;
			if (i%prime[j]==0) break;
		}
	}
}
signed main() {
	getPrime(4e7);
	int t;
	cin>>t;
	while (t--) {
		int n;
		cin>>n;
		int cnt = 0;
		int l=1,r=1;
		int sum = 2;
		while (r<=tot&&prime[r]<=n) {
			while (sum>=n) {
				if (sum==n) cnt++;
				l++;
				sum -= prime[l-1];
			}
			while (sum<n) {
				r++;
				sum += prime[r];
			}
		}
		cout<<cnt<<'\n';
	}
}

　　E-X-factor Chains_牛客競賽數學專題班整數分解與篩法 (nowcoder.com)

對x分解質因數後，最大值就是每個質數因子的指數和，方案書用雜湊+記憶化搜尋+遞迴分治求出

#include <bits/stdc++.h>
#define fi first
#define se second
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const int maxn = 1e6+7;
const int LOG = 32;
const ll base = 127;
const ll p = 100029383;
int a[LOG];
int vis[maxn];
int prime[maxn];
int tot = 0;
int n;
map<ll,ll> dp;
ll getHash() {
	ll sum = 0;
	for (int i=1;i<=n;i++) {
		sum = (sum*base+a[i])%p;
	}
	return sum;
}
void getPrime(int n) {
	for (int i=2;i<=n;i++) {
		if (!vis[i]) prime[++tot] = i;
		for (int j=1;j<=tot;j++) {
			if (i*prime[j]>n) break;
			vis[i*prime[j]] = 1;
			if (i%prime[j]==0) break;
		}
	}
}
int getFac(int x) {
	int ind = 1;
	int j = 1;
	if (!vis[x]) {
		a[ind] = 1;
		return ind;
	}
	while (x>1) {
		if (!vis[x]) {
			a[ind]++;
			ind++;
			break;
		}
		while (x%prime[j]!=0) j++;
		while (x%prime[j]==0) {
			a[ind]++;
			x /= prime[j];
		}
		ind++;	
	}
	return ind-1;

	
}
ll func() {
	int cnt = 0;
	for (int i=1;i<=n;i++) {
		if (a[i]>0) cnt++;
	}
	if (cnt==1) return 1;
	ll h = getHash();
	if (dp[h]) return dp[h];
	ll res = 0;
	for (int i=1;i<=n;i++) {
		if (a[i]==0) continue;
		a[i]--;
		res += func();
		a[i]++;
	}
	return dp[h] = res;
}
signed main() {
	IOS;
	int t;
	cin>>t;
	getPrime(maxn-1);
	while (t--) {
		memset(a,0,sizeof a);
		int x;
		cin>>x;
		n = getFac(x);
		int sum = 0;
		for (int i=1;i<=n;i++) {
			sum += a[i];
		}
		sort(a+1,a+1+n);
		cout<<sum<<" "<<func()<<'\n';
	}
}

　　97. 約數之和 - AcWing題庫

對A質因數分解，由正約數和公式，並且該公式的每個因子都是個等比數列和，可以直接算出結果，並且特判下乘法逆元不存在的情況

#include <bits/stdc++.h>
#define fi first
#define se second
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define L k<<1
#define R k<<1|1
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const int maxn = 5e7+7;
const ll p = 9901;
struct Node {
	int p,a;
	Node() {
		a = 0;
	}
};
ll quick(ll x,ll n) {
	ll res = 1;
	while (n) {
		if (n&1) res = res * x % p;
		x = x * x % p;
		n >>= 1;
	}
	return res;
}
int prime[int(1e4+7)];
bool vis[int(1e4+7)];
int tot = 0;
Node fac[100];
void getPrime(int n) {
	for (int i=2;i<=n;i++) {
		if (!vis[i]) prime[++tot] = i;
		for (int j=1;j<=n;j++) {
			if (i*prime[j]>n) break;
			vis[i*prime[j]] = 1;
			if (i%prime[j]==0) break;
		}
	}
}
int getFac(int x) {
	int ind = 1;
	int j = 1;
	while (x>1) {
		while (x%prime[j]!=0) {
			j++;
			if (prime[j]*prime[j]>x) {
				fac[ind].p = x;
				fac[ind].a = 1;
				ind++;
				return ind-1;
			}
		}
		while (x%prime[j]==0) {
			fac[ind].p = prime[j];
			fac[ind].a++;
			x /= prime[j];
		}
		ind++;
		
	}
	return ind-1;
}
int main() {
	int A,B;
	cin>>A>>B;
	if (A==0) 
	{
		cout<<0;
		return 0;
	}
	getPrime(1e4);
	int n = getFac(A);
	ll res = 1;
	for (int i=1;i<=n;i++) {
		if ((fac[i].p-1)%p!=0) res = ((res*(quick(fac[i].p,fac[i].a*B+1)-1)%p*quick(fac[i].p-1,p-2))%p+p)%p;
		else res = res*(fac[i].a*B+1)%p;
	}
	cout<<res;
}

A New Function | LightOJ

易推出，由於n的範圍大，故要數論分塊，可得出答案

#include <bits/stdc++.h>
#define fi first
#define se second
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define L k<<1
#define R k<<1|1
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
const int maxn = 5e7+7;
int main() {
	int t;
	cin>>t;
	int cnt = 0;
	while (t--) {
		cnt++;
		ll n;
		cin>>n;
		ll r;
		ll res = 0;
		for (ll l=2;l<=n;l=r+1) {
			r = n/(n/l);
			res += (l+r)*(r-l+1)/2*(n/l-1);
		}
//		res -= (n+2)*(n-1)/2;
		cout<<"Case "<<cnt<<": ";
		cout<<res<<'\n';
	}
}