数学相关模板-基础

反素数

求1-n以内约数最多的数

#include <iostream>

int p[16] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 };
unsigned long long n;
unsigned long long ans,
ans_num;  // ans 为 n 以内的最大反素数（会持续更新），ans_sum 为
// ans的因子数。

// depth: 当前在枚举第几个素数
// temp: 当前因子数量为 num的时候的数值
// num: 当前因子数
// up：上一个素数的幂，这次应该小于等于这个幂次嘛
void dfs(int depth, unsigned long long temp, unsigned long long num, int up) {
    if (depth >= 16 || temp > n) return;
    if (num > ans_num) {  // 更新答案
        ans = temp;
        ans_num = num;
    }
    if (num == ans_num && ans > temp) ans = temp;  // 更新答案
    for (int i = 1; i <= up; i++) {
        if (temp * p[depth] > n)
            break;  // 剪枝：如果加一个这个乘数的结果比ans要大，则必不是最佳方案
        dfs(depth + 1, temp *= p[depth], num * (i + 1),
            i);  // 取一个该乘数，进行对下一个乘数的搜索
    }
    return;
}

using std::cin;
using std::cout;

int main() {
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
    cin >> n;

    ans_num = 0;
    dfs(0, 1, 1, 60);
    cout << ans << " " << ans_num;

    return 0;
}

(数，对应的约数个数)

1e5, 83160 128
1e7, 8648640 448
1e9, 735134400 1344
1e12, 963761198400 6720

[ABC248G] GCD cost on the tree

题目描述

给定一颗树有 $n$ 个结点，每个结点上有一个权值 $a_{i}$ , 对于每条至少包含两个点的简单路径，它的贡献为路径上点的数量(包括端点) $\times$ 路径上所有点的 $a_{i}$
的最大公约数(gcd)。
求所有简单路径的贡献之和，对 $998244353$ 取模。

假设当前计算gcd为d的所有路径和长度，直接计算不好计算，我们可以计算所有d|x的路径长度和，然后再减掉d的倍数的影响即可。

因为1e5以内，约数最多的数只有128，因此每个数最多只会加入128次，那么我们对于每个d，我们完全可以将树建出来，然后利用统计sz计算每个点被经过多少次。

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define eb emplace_back
#define pi pair<ll,ll>
#define mk(x,y) make_pair((x),(y))
#define lc (o<<1)
#define rc ((o<<1)|1)
#define fi first
#define se second
using namespace std;
typedef long long ll;
const ll inf = 1ll << 60;
const int N = 5e5 + 10;
const ll mo = 998244353;
vector<int> tr[N];
ll n, a[N], x, y, f[N];
ll p[N], tot, v[N], num;
bool vis[N];
vector<int> t[N], d;
queue<int> q;
vector<int> e[N];
bool bz[N], in[N];
ll sz[N], g[N], sum, c[N], u[N], ans;
void work(int x, int pos) {
    d.clear();
    d.eb(1);
    int z = v[x], l, r;
    while (x != 1) {
        z = v[x];
        l = 0; r = d.size();
        while (x % z == 0) {
            x /= z;
            fo(i, l, r - 1) d.eb(d[i] * z);
            l = r; r = d.size();
        }
    }

    for (auto x : d) t[x].eb(pos);
}
void ins(int x) {
    q.push(x);
    bz[x] = 1;
    work(a[x], x);
}
void bfs() {
    ins(1);

    while (!q.empty()) {
        int x = q.front(); q.pop();
        for (int v : tr[x]) {
            if (!bz[v]) {
                f[v] = x;
                ins(v);
            }
        }
    }
}
void add(ll& x, ll y) {
    x = (x + y) % mo;
}
void dg(int x) {
    add(num, (sum - sz[x]) * sz[x] % mo + g[x]);
    for (int v : e[x]) dg(v);
}
void dfs(int x) {
    sz[x] = 0;
    ll s = 0;
    for (int v : e[x]) {
        dfs(v);
        sz[x] += sz[v];
        s += sz[v] * sz[v];
    }

    g[x] = sz[x] + (sz[x] * sz[x] - s) / 2;
    sz[x]++;
}
void solve()
{
    fo(i, 2, N - 1) {
        if (!vis[i]) {
            p[++tot] = i;
            v[i] = i;
        }
        fo(j, 1, tot) {
            if (i * p[j] >= N) break;
            vis[i * p[j]] = 1;
            if (i % p[j] == 0) {
                v[i * p[j]] = p[j];
                break;
            }
            else v[i * p[j]] = p[j];
        }
    }

    cin >> n;
    fo(i, 1, n) cin >> a[i];

    fo(i, 1, n - 1) {
        cin >> x >> y;
        tr[x].eb(y);
        tr[y].eb(x);
    }

    bfs();

    // for (auto x : t[1]) cout << x << "\n";
    // cout << "\n";
    // fo(i, 1, n)cout << f[i] << "\n";

    fd(i, 1e5, 1) {
        for (auto x : t[i]) {
            in[x] = 1;
            if (in[f[x]]) e[f[x]].eb(x);
        }

        ll z = 0;
        for (auto x : t[i]) {
            if (!in[f[x]]) {
                dfs(x);
                sum = sz[x];
                num = 0;
                dg(x);
                add(z, num);
            }
        }

        fo(j, 2, 1e5 / i) {
            z = (z - u[i * j]) % mo;
        }
        u[i] = z;
        add(ans, z* i% mo);

        for (auto x : t[i]) {
            e[x].clear();
            in[x] = 0;
        }
    }

    ans = (ans % mo + mo) % mo;
    cout << ans;
}
int main()
{
    // freopen("data.in", "r", stdin);
    // freopen("data.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

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

欧拉定理

$(a, m) = 1, a^{φ (m)} \equiv 1 (\mod m)$

扩展欧拉定理

当 $(a, m) \neq 1$ 时，

$b < φ (m), a^{b} \equiv a^{b} (\mod m)$
$b \geq φ (m), a^{b} \equiv a^{b m o d φ (m) + φ (m)} (\mod m)$

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mk(x,y) make_pair((x),(y))
#define eb emplace_back
#define A puts("Yes")
#define B puts("No")
#define fi first
#define se second
#define pi pair<ll,ll>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef long long i64;
//typedef __int128 i128;
const int inf=1e9;
const ll mo1=1e9+9;
const ll mo2=1e9+7;
const ll P=131;
const ll Q=13331;
//const ll mo=1e9+7;
const int N=1e5+5;
char ch;
ll a,m,b;
ll power(ll a,ll b){
	ll t=1,y=a;
	while (b) {
		if (b&1) t=t*y%m;
		y=y*y%m;
		b/=2;
	}
	return t;
}
int main(){
//	freopen("data.in","r",stdin);
//	freopen("data.out","w",stdout);

	scanf("%lld %lld",&a,&m);
	a%=m;
	
	ll t=m,phi=m;
	for (ll i=2;i*i<=m;i++) {
		if (t%i==0) {
			phi=phi/i*(i-1);
			while (t%i==0) t/=i;
		}
	}
	if (t!=1) phi=phi/t*(t-1);
	
	bool flag=0;
	char ch;
	for (ch=getchar();!('0'<=ch && ch<='9');ch=getchar());
	for (;('0'<=ch && ch<='9');ch=getchar()) {
		b=b*10+ch-'0';
		if (b>=phi) {
			b%=phi;
			flag=1;
		}
	}
	
	if (flag) printf("%lld\n",power(a,b+phi));
	else printf("%lld\n",power(a,b));
		
	
	return 0;
}

质因数分解

$m [x]$ 表示x的最小质因数

void work(int x){
	e.clear();
	e.push_back(1);
	while (x!=1){
		z=m[x];
		
		l=0; r=e.size();
		while (x%z==0) {
			x/=z;
			fo(i,l,r-1) e.push_back(e[i]*z);
			l=r; r=e.size();
		}
	}
}

Pollard-Rho 算法

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

int t;
long long max_factor, n;

long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}

long long quick_pow(long long x, long long p, long long mod) {  // 快速幂
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (__int128)ans * x % mod;
    x = (__int128)x * x % mod;
    p >>= 1;
  }
  return ans;
}

bool Miller_Rabin(long long p) {  // 判断素数
  if (p < 2) return 0;
  if (p == 2) return 1;
  if (p == 3) return 1;
  long long d = p - 1, r = 0;
  while (!(d & 1)) ++r, d >>= 1;  // 将d处理为奇数
  for (long long k = 0; k < 10; ++k) {
    long long a = rand() % (p - 2) + 2;
    long long x = quick_pow(a, d, p);
    if (x == 1 || x == p - 1) continue;
    for (int i = 0; i < r - 1; ++i) {
      x = (__int128)x * x % p;
      if (x == p - 1) break;
    }
    if (x != p - 1) return 0;
  }
  return 1;
}

long long Pollard_Rho(long long x) {
  long long s = 0, t = 0;
  long long c = (long long)rand() % (x - 1) + 1;
  int step = 0, goal = 1;
  long long val = 1;
  for (goal = 1;; goal *= 2, s = t, val = 1) {  // 倍增优化
    for (step = 1; step <= goal; ++step) {
      t = ((__int128)t * t + c) % x;
      val = (__int128)val * abs(t - s) % x;
      if ((step % 127) == 0) {
        long long d = gcd(val, x);
        if (d > 1) return d;
      }
    }
    long long d = gcd(val, x);
    if (d > 1) return d;
  }
}

void fac(long long x) {
  if (x <= max_factor || x < 2) return;
  if (Miller_Rabin(x)) {              // 如果x为质数
    max_factor = max(max_factor, x);  // 更新答案
    return;
  }
  long long p = x;
  while (p >= x) p = Pollard_Rho(x);  // 使用该算法
  while ((x % p) == 0) x /= p;
  fac(x), fac(p);  // 继续向下分解x和p
}

int main() {
  scanf("%d", &t);
  while (t--) {
    srand((unsigned)time(NULL));
    max_factor = 0;
    scanf("%lld", &n);
    fac(n);
    if (max_factor == n)  // 最大的质因数即自己
      printf("Prime\n");
    else
      printf("%lld\n", max_factor);
  }
  return 0;
}

fwt相关

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mk(x,y) make_pair((x),(y))
#define eb emplace_back
#define A puts("Yes")
#define B puts("No")
#define fi first
#define se second
#define pi pair<ll,ll>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef long long i64;
typedef unsigned int uint;
//typedef __int128 i128;
const int inf = 1e9;
const ll mo1 = 1e9 + 9;
const ll mo2 = 1e9 + 7;
const ll P = 131;
const ll Q = 13331;
const ll mo = 998244353;
const int N = 1 << 20;
ll inv2, a[N], b[N], n, aa[N], bb[N], c[N];
void add(ll& x, ll y) {
    x = (x + y) % mo;
}
void Or(ll* a, ll type) {
    for (ll x = 2;x <= n;x <<= 1) {
        ll k = x >> 1;
        for (ll i = 0;i < n;i += x) {
            for (ll j = 0;j < k;j++) {
                add(a[i + j + k], a[i + j] * type);
            }
        }
    }
}
void And(ll* a, ll type) {
    for (ll x = 2;x <= n;x <<= 1) {
        ll k = x >> 1;
        for (ll i = 0;i < n;i += x) {
            for (ll j = 0;j < k;j++) {
                add(a[i + j], a[i + j + k] * type);
            }
        }
    }
}
void Xor(ll* a, ll type) {
    for (ll x = 2;x <= n;x <<= 1) {
        ll k = x >> 1;
        for (ll i = 0;i < n;i += x) {
            for (ll j = 0;j < k;j++) {
                add(a[i + j], a[i + j + k] % mo);
                a[i + j + k] = (a[i + j] - 2ll * a[i + j + k] % mo) % mo;
                a[i + j] = a[i + j] * type % mo;
                a[i + j + k] = a[i + j + k] * type % mo;
            }
        }
    }
}
int main() {
    //	freopen("data.in","r",stdin);
    //	freopen("data.out","w",stdout);

    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);

    inv2 = -(mo / 2);
    inv2 = (inv2 % mo + mo) % mo;

    cin >> n;
    n = 1 << n;

    fo(i, 0, n - 1) cin >> a[i];
    fo(i, 0, n - 1) cin >> b[i];

    fo(i, 0, n - 1) aa[i] = a[i], bb[i] = b[i];
    Or(aa, 1); Or(bb, 1);
    fo(i, 0, n - 1) c[i] = aa[i] * bb[i] % mo;
    Or(c, -1);
    fo(i, 0, n - 1) cout << (c[i] % mo + mo) % mo << " ";
    cout << "\n";

    fo(i, 0, n - 1) aa[i] = a[i], bb[i] = b[i];
    And(aa, 1); And(bb, 1);
    fo(i, 0, n - 1) c[i] = aa[i] * bb[i] % mo;
    And(c, -1);
    fo(i, 0, n - 1) cout << (c[i] % mo + mo) % mo << " ";
    cout << "\n";

    fo(i, 0, n - 1) aa[i] = a[i], bb[i] = b[i];
    Xor(aa, 1); Xor(bb, 1);
    fo(i, 0, n - 1) c[i] = aa[i] * bb[i] % mo;
    Xor(c, inv2);
    fo(i, 0, n - 1) cout << (c[i] % mo + mo) % mo << " ";
    cout << "\n";




    return 0;
}

[TJOI2011] 01矩阵

题目描述

$n \times m$ 的 $01$ 矩阵，其中某些位置已经确定，为 '.' 的位置可以填 $0$ 或 $1$ ，求相邻两个位置不同为 $1$ 的矩阵方案数，答案模 $10007$ 。

首先预处理出哪些状态是合法的，假设这一行。的状态是s，那么能够由上一行转移来的t必定满足s|t=0,因此t是s补集中的子集，直接利用or卷积即可。

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mk(x,y) make_pair((x),(y))
#define eb emplace_back
#define A puts("Yes")
#define B puts("No")
#define fi first
#define se second
#define pi pair<ll,ll>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef long long i64;
typedef unsigned int uint;
//typedef __int128 i128;
const int inf = 1e9;
const ll mo1 = 1e9 + 9;
const ll mo2 = 1e9 + 7;
const ll P = 131;
const ll Q = 13331;
const ll mo = 10007;
const int N = 1 << 20;
int c[300][300];
ll inv2, a[N], b[N], n, aa[N], bb[N], m, f[N], g[N], st[N], one[N], zero[N];
string s;
bool bz[N];
void add(ll& x, ll y) {
    x = (x + y) % mo;
}
void Or(ll* a, ll type) {
    for (ll x = 2;x <= n;x <<= 1) {
        ll k = x >> 1;
        for (ll i = 0;i < n;i += x) {
            for (ll j = 0;j < k;j++) {
                add(a[i + j + k], a[i + j] * type);
            }
        }
    }
}
void And(ll* a, ll type) {
    for (ll x = 2;x <= n;x <<= 1) {
        ll k = x >> 1;
        for (ll i = 0;i < n;i += x) {
            for (ll j = 0;j < k;j++) {
                add(a[i + j], a[i + j + k] * type);
            }
        }
    }
}
void Xor(ll* a, ll type) {
    for (ll x = 2;x <= n;x <<= 1) {
        ll k = x >> 1;
        for (ll i = 0;i < n;i += x) {
            for (ll j = 0;j < k;j++) {
                add(a[i + j], a[i + j + k] % mo);
                a[i + j + k] = (a[i + j] - 2ll * a[i + j + k] % mo) % mo;
                a[i + j] = a[i + j] * type % mo;
                a[i + j + k] = a[i + j + k] * type % mo;
            }
        }
    }
}
int main() {
    //	freopen("data.in","r",stdin);
    //	freopen("data.out","w",stdout);

    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);

    inv2 = -(mo / 2);
    inv2 = (inv2 % mo + mo) % mo;

    cin >> n >> m;
    fo(i, 1, n) {
        cin >> s;
        fo(j, 1, m) {
            if (s[j - 1] == '.') c[i][j] = -1;
            else c[i][j] = s[j - 1] - '0';
        }
    }

    if (n > 15) {
        fo(i, 1, n) fo(j, i + 1, n) swap(c[i][j], c[j][i]);
        swap(n, m);
    }

    fo(j, 1, m) {
        fo(i, 0, n - 1) {
            if (c[i + 1][j] == 0) zero[j] |= (1 << i);
            if (c[i + 1][j] == 1) one[j] |= (1 << i);
        }
    }

    int k = n;

    n = 1 << n;
    fo(s, 0, n - 1) {
        fo(i, 0, k - 2) {
            if ((s & (1 << i)) && (s & (1 << (i + 1)))) bz[s] = 1;
        }
    }

    f[0] = 1;
    fo(i, 1, m) {
        Or(f, 1);
        fo(j, 0, n - 1) {
            if (bz[j] || (j & zero[i]) || ((j & one[i]) != one[i])) g[j] = 0;
            else g[j] = f[j ^ (n - 1)];
        }
        fo(j, 0, n - 1) f[j] = g[j];
    }

    ll ans = 0;
    fo(j, 0, n - 1) add(ans, g[j]);
    ans = (ans % mo + mo) % mo;
    cout << ans;


    return 0;
}

[HAOI2015] 按位或

题目描述

刚开始你有一个数字 $0$ ，每一秒钟你会随机选择一个 $[0, 2^{n} - 1]$ 的数字，与你手上的数字进行或（C++,C 的 |，pascal 的 or）操作。选择数字 $i$ 的概率是 $p_{i}$ 。保证 $0 \leq p_{i} \leq 1$ ， $\sum p_{i} = 1$ 。问期望多少秒后，你手上的数字变成 $2^{n} - 1$ 。

需要利用min-max容斥， $E (m a x (S)) = \sum_{T \in S} (- 1)^{| T | + 1} E (m i n (T))$
考虑怎么求E(min(T))，假设选中T集合中的位置的概率为p，则 $E (m i n (T)) = \frac{1}{p}$ ，而p的计算为所有至少包含T中一个位的所有pi的和，那么我们减去跟t完全没有交集的集合的pi的和即可，也就是补集的子集，最后计算min-max容斥的式子再用一次计算子集的套路即可。

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mk(x,y) make_pair((x),(y))
#define pi pair<ll,ll>
#define eb emplace_back
#define A puts("Yes")
#define B puts("No")
#define fi first
#define se second
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef long long i64;
const ll inf = 1ll << 60;
const ll mo1 = 1e9 + 9;
const ll mo2 = 1e9 + 7;
const ll P = 131;
const ll Q = 13331;
const ll mo = 998244353;
const int N = 1 << 20;
const db eps = 1e-9;
ll n, st;
db a[N], p[N];
int main() {
	// freopen("data.in", "r", stdin);
	// freopen("data.out", "w", stdout);

	ios::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);

	db sum = 0;
	cin >> n;
	int st = (1 << n) - 1;
	fo(i, 0, st) cin >> p[i], sum += p[i];

	fo(j, 0, n - 1) fo(i, 0, st) {
		if ((i >> j) & 1) p[i] += p[i ^ (1 << j)];
	}

	fo(i, 0, st) a[i] = sum - p[st ^ i];

	bool flag = 1;
	fo(i, 1, st) if (a[i] < eps) flag = 0;

	if (!flag) {
		cout << "INF";
		return 0;
	}

	a[0] = 0;
	fo(i, 1, st) {
		int cnt = 0;
		fo(j, 0, n - 1) if ((i >> j) & 1)  ++cnt;
		a[i] = 1.0 / a[i];
		if (cnt % 2 == 0) a[i] = -a[i];
	}

	fo(j, 0, n - 1) fo(i, 0, st) {
		if ((i >> j) & 1) a[i] += a[i ^ (1 << j)];
	}

	cout << fixed << setprecision(10) << a[st];



	return 0;

}

excrt

【模板】扩展中国剩余定理（EXCRT）

题目描述

给定 $n$ 组非负整数 $a_{i}, b_{i}$ ，求解关于 $x$ 的方程组的最小非负整数解。

{\begin{cases} x \equiv b_{1} (\mod a_{1}) \\ x \equiv b_{2} (\mod a_{2}) \\ \dots \\ x \equiv b_{n} (\mod a_{n}) \end{cases}

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define lc (o<<1)
#define rc ((o<<1)|1)
#define mk(x,y) make_pair((x),(y))
#define eb emplace_back
#define A puts("Yes")
#define B puts("No")
#define fi first
#define se second
#define pi pair<ll,ll>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef long long i64;
typedef unsigned int uint;
//typedef __int128 i128;
const int inf = 1e9;
const ll mo1 = 1e9 + 9;
const ll mo2 = 1e9 + 7;
const ll P = 131;
const ll Q = 13331;
//const ll mo=1e9+7;
const int N = 2e5 + 5;
struct node {
    ll r, m;
    bool flag;
};
node a[N];
ll n;
ll gcd(ll a, ll b) {
    if (!b) return a;
    return gcd(b, a % b);
}
ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}
void exgcd(ll a, ll b, ll& d, ll& x, ll& y) {
    if (!b) {
        d = a;
        x = 1;
        y = 0;
        return;
    }
    exgcd(b, a % b, d, y, x);
    y -= (a / b) * x;
}
ll mul(ll a, ll b, ll mod) {
    a %= mod;
    b %= mod;
    ll t = 0;
    while (b) {
        if (b & 1) t = (t + a) % mod;
        a = (a + a) % mod;
        b /= 2;
    }
    return t;
}
node merge(node a, node b) {
    ll r1, r2, m1, m2, x, y, d, mm, rr;

    if (!a.flag) return (node) { 0, 0, 0 };
    r1 = a.r; m1 = a.m;
    r2 = b.r; m2 = b.m;
    exgcd(m1, -m2, d, x, y);
    if ((r2 - r1) % d) return (node) { 0, 0, 0 };

    mm = lcm(m1, m2);
    x = mul((r2 - r1) / d, x, mm);

    rr = (mul(x, m1, mm) + r1 % mm) % mm;
    return (node) { rr, mm, 1 };
}
int main() {
    //	freopen("data.in","r",stdin);
    //	freopen("data.out","w",stdout);

    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);

    cin >> n;
    fo(i, 1, n) cin >> a[i].m >> a[i].r, a[i].flag = 1;

    node ans = a[1];
    fo(i, 2, n) {
        ans = merge(ans, a[i]);
    }

    ll r = ans.r, m = ans.m;
    cout << (r % m + m) % m;

    return 0;
}

矩阵乘法

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define eb emplace_back
using namespace std;
typedef long long ll;
const ll mo=998244353;
struct mat{
	ll a[6][6];
	mat(){
		memset(a,0,sizeof(a));
	}
};
mat operator* (mat a,mat b){
	mat c;
	fo(i,0,5) fo(j,0,5) fo(k,0,5) {
		add(c.a[i][j], a.a[i][k]*b.a[k][j]%mo);
	}
	return c;
}
void power(ll b){
	while (b) {
		if (b&1) {
			t=y*t;
		}
		y=y*y;
		b/=2;
	}
}

【模板】Dirichlet 前缀和

题目背景

模板题，无背景。

题目描述

给定一个长度为 $n$ 的数列 $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ 。

现在你要求出一个长度为 $n$ 的数列 $b_{1}, b_{2}, b_{3}, \dots, b_{n}$ ，满足

b_{k} = \sum_{i | k} a_{i}

由于某些神秘原因，这里的 $b_{k}$ 要对 $2^{32}$ 取模。
类似于高维前缀和

#include<bits/stdc++.h>
#define fo(i,a,b) for (ll (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (ll (i)=(b);(i)>=(a);(i)--)
#define eb emplace_back
#define pi pair<ll,ll>
#define mk(x,y) make_pair((x),(y))
#define lc (o<<1)
#define rc ((o<<1)|1)
//#define A puts("Yes")
//#define B puts("No")
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef double db;
const ll mo1=1e9+7;
const ll mo2=1e9+9;
const ll P=131;
const ll Q=13331;
const ll inf=1e9;
const int N=2e7+5;
#define uint unsigned int
uint seed;
uint a[N],ans;
bool bz[N];
int n;
inline uint getnext(){
	seed^=seed<<13;
	seed^=seed>>17;
	seed^=seed<<5;
	return seed;
}
int main() {
//    freopen("data.in", "r", stdin);
//    freopen("data.out", "w", stdout);
	
	cin>>n>>seed;
	fo(i,1,n) a[i]=getnext();
	
	ans=a[1];
	for (int i=2;i<=n;i++) {
		if (!bz[i]) {
			for (int j=2;j<=n/i;j++) {
				bz[i*j]=1;
				a[i*j]+=a[j];
			}
		}
		ans^=a[i]+a[1];
	}
	cout<<ans;
	
    return 0;
    


}

posted @ 2024-10-18 10:08 gan_coder 阅读(7) 评论(0) 编辑收藏举报

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