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蓝桥杯 6.数学

线性代数与矩阵运算&数论

矩阵乘法&整除&同余&GCD&LCM

矩阵乘法

介绍

• 线性代数中的基础内容. 只有当相乘的两个矩阵的左矩阵的列数和右矩阵的行数相同时, 才能相乘
• $n\ast m$的矩阵和$m\ast k$的矩阵做乘法后的矩阵大小为$n\ast k$

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e3 + 5;
ll a[N][N], b[N][N];int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll n, m, k; cin >> n >> m >> k;for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){cin >> a[i][j];}}for(int i = 1; i <= m; i++){for(int j = 1; j <= k; j++){cin >> b[i][j];}}for(int i = 1; i <= n; i++){for(int j = 1; j <= k; j++){ll ans = 0;for(int x = 1; x <= m; x++){ans += a[i][x] * b[x][j];}cout << ans << " ";}cout << "\n";}return 0;
}

整除

简介

• 在计算机中, 整数之间的除法往往是整除且向下取整
• $12/5=2$如果想要向上取整$12/5=3$, 只需要将式子改为$(a+b-1)/b$, 即$(12+5-1)/5$
• 如果x可以被y整除, 记作$y|x$, 例如$4|12$
• 整除的性质

同余

简介

• 两个或多个数字x, 对于一个模数M的余数是相等的, 或者说在模M意义下, 他们是相等的
• 例如M=7, x=[1, 8, 15, 22, 50]是同余的, 这些x都有一个共同的性质, 就是x%M=1

GCD&LCM

简介

• GCD是最大公约数, LCM是最小公倍数, 大多数情况下, 更关注GCD, 而LCM可以通过GCD求出来
• GCD通过欧几里得辗转相除法得到, 不必深究其原理, 学会使用即可
• C++已经准备好了gcd和lcm函数, 可以直接调用, 分别为: __gcd(a, b), __lcm(a, b)

int gcd(int a, int b){return b == 0 ? a : gcd(b, a % b);
}

int lcm(int a, int b){return a / gcd(a, b) * b;
} // 这里推荐先做除法, 再做乘法, 避免溢出

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e3 + 5;
ll n, m;ll gcd(int a, int b){while(b){int digit = a % b;a = b;b = digit;}return a;
}void AC(){cin >> n >> m;// cout << __gcd(n, m) << "\n";cout << gcd(n, m) << "\n";
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll t; cin >> t;while(t--){AC();}return 0;
}

扩展欧几里得

算法原理

• 扩展欧几里得不仅可以求到gcd, 而且还可以求解到不定方程$ax+by=c$的解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;ll exgcd(ll a, ll b, ll &x, ll &y){if(!b){// 递归出口x = 1, y = 0;return a;}// 注意这里交换了x, y的参数位置// 那么得到的y就已经是下一层的x了, 而y也只需要减去a/b*x即可ll d = exgcd(b, a % b, y, x);y -= a / b * x;return d;
}

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 2e5 + 5;
ll a, b, m;ll exgcd(ll a, ll b, ll &x, ll &y){if(!b){// 递归出口x = 1, y = 0;return a;}// 注意这里交换了x, y的参数位置// 那么得到的y就已经是下一层的x了, 而y也只需要减去a/b*x即可ll d = exgcd(b, a % b, y, x);y -= a / b * x;return d;
}ll getabs(ll x) {return x < 0 ? -x : x;
}void AC(){cin >> a >> b >> m;ll x, y, d = exgcd(getabs(a), getabs(m), x, y);if(b % d){cout << -1 << "\n";} else {x *= (a < 0 ? -1 : 1) * (b / d);x = (x % (m / d) + (m / d)) % (m / d);cout << x << "\n";}
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll t; cin >> t;while(t--){AC();}return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
// const ll N = 2e5 + 5;
ll _x, _y, m, n, l;ll exgcd(ll a, ll b, ll &x, ll &y){if(!b){// 递归出口x = 1, y = 0;return a;}// 注意这里交换了x, y的参数位置// 那么得到的y就已经是下一层的x了, 而y也只需要减去a/b*x即可ll d = exgcd(b, a % b, y, x);y -= a / b * x;return d;
}ll getabs(ll x) {return x < 0 ? -x : x;
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cin >> _x >> _y >> m >> n >> l;ll a = m - n, b = l, c = _y - _x;ll x, y, d = exgcd(getabs(a), getabs(b), x, y);if(c % d){cout << "impossible" << "\n";} else {x *= (a < 0 ? -1 : 1) * (c / d);x = (x % (b / d) + (b / d)) % (b / d);cout << x << "\n";}return 0;
}

编程156-158

156-157没写

// 这个题不能使用递归记忆化来写, 应该使用矩阵快速幂求解
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 5;
const ll p = 1e9 + 7;
ll n, k, a[N];struct Matrix{ll val[N][N];Matrix(){memset(val, 0, sizeof val);}
};Matrix mul(Matrix &a, Matrix &b){Matrix c;for(int i = 1; i <= 2; i ++){for(int j =  1; j <= 2; j++){for(int k = 1; k <= 2; k++){c.val[i][j] = (c.val[i][j] + a.val[i][k] * b.val[k][j]) % p;}}}return c;
}Matrix fastPow(Matrix a, ll b){Matrix ret;for(int i = 1; i <= 2; i++){ret.val[i][i] = 1;}while(b){if(b & 1) ret = mul(ret, a);a = mul(a, a);b >>= 1;}return ret;
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll t; cin >> t;while(t--){cin >> k;if(k == 1 || k == 2) cout << 1 << "\n";else {Matrix xs;xs.val[1][1] = 1;xs.val[1][2] = 1;xs.val[2][1] = 1;Matrix ans = fastPow(xs, k - 2);cout << (ans.val[1][1] + ans.val[1][2]) % p << "\n";}}return 0;
}

素数朴素判定&埃氏筛法

朴素的素数判定法

简介

• 又称质数, 指的是除了1和它本身, 没有其他的因子的数字

bool isPrime(int n){if(n < 2) return false; // <2的一定不是质数for(int i = 2; i <= n / i; i++){if(n % i == 0) return false; // i<=n/i等价于i*i<=n, 这样可以防止溢出(sqrt()函数速度会比较慢, 而且返回类型是浮点型)}return true;
}

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 5;
const ll p = 1e9 + 7;bool isPrime(ll n){if(n < 2) return false; // <2的一定不是质数for(int i = 2; i <= n / i; i++){if(n % i == 0) return false; // i<=n/i等价于i*i<=n, 这样可以防止溢出(sqrt()函数速度会比较慢, 而且返回类型是浮点型)}return true;
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll n; cin >> n;ll ans = 0;for(int i = 1; i <= n; i++){ll sum = 0, j = i;while(j){sum += j % 10;j /= 10;}if(isPrime(sum)){ans++;}}cout << ans << "\n";return 0;
}

埃氏筛法

简介

• 对于一个合数(即不是素数), 它必然存在一个质数的因子, 也就是说合数一定是某个较小的质数的倍数, 所以我们不妨枚举倍数
• 埃氏筛法: 已知2是一个质数, 然后将2的所有大于它本身的倍数全部"筛掉", 因为可以判断它们都不是质数, 然后到3, 4…, 如果枚举到的数字i没有被筛掉, 说明在[2~i-1]不存在某个数字可以整除i, 于是就可以直接判定i是一个质数, 然后再将i所有的倍数全部筛掉

bool vis[N]; // vis[i] = true表示被筛掉了
vis[0] = vis[1] = true;
for(int i = 2; i <= n; i++){if(!vis[i]){ // 没有被筛掉for(int j = 2 * i; j <= n; j += i){ // 枚举所有大于i的倍数, 也就是从2i开始, j+=i也就是i的倍数vis[j] = true; // 筛掉}}
}

例题讲解

// 暴力枚举
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5 + 5;
const ll p = 1e9 + 7;
vector<ll> vec;
bool isPrime(ll n){if(n < 2) return false; // <2的一定不是质数for(int i = 2; i <= n / i; i++){if(n % i == 0) return false; // i<=n/i等价于i*i<=n, 这样可以防止溢出(sqrt()函数速度会比较慢, 而且返回类型是浮点型)}return true;
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll n; cin >> n;for(int i = 1; i <= n; i++){if(isPrime(i)){vec.push_back(i);}}ll ans = 0;for(int i = 0; i < vec.size(); i++){for(int j = i; j < vec.size(); j++){if(isPrime(vec[j] - vec[i])){ans++;}}}cout << ans << "\n";return 0;
}// 埃氏筛法
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5 + 5;
const ll p = 1e9 + 7;
vector<ll> vec;
bool vis[N]; // vis[i] = true表示被筛掉了void init(ll n){vis[0] = vis[1] = true;for(int i = 2; i <= n; i++){if(!vis[i]){ // 没有被筛掉for(int j = 2 * i; j <= n; j += i){ // 枚举所有大于i的倍数, 也就是从2i开始, j+=i也就是i的倍数vis[j] = true; // 筛掉}}}for(int i = 2; i <= n; i++){if(!vis[i]){vec.push_back(i);}}
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll n; cin >> n;init(n);ll ans = 0;for(int i = 0; i < vec.size(); i++){for(int j = i; j < vec.size(); j++){if(!vis[vec[j] - vec[i]]){ans++;}}}cout << ans << "\n";return 0;
}

欧拉筛

原理

• 在埃氏筛法中, 枚举每个素数并将其所有的倍数筛除, 其中是存在一些冗余的, 例如数字30会被2筛除, 也会被3和5筛除. 如果可以保证每个合数只筛除一次, 那么就可以降低时间复杂度, 欧拉筛就是做了这么一件事, 让素数筛法的时间复杂度降低到了O(n)

void euler(int n){vector<int> primes;vis[0] = vis[1] = true;for(int i = 2; i <= n; i++){// 如果i没有被筛除, 说明i是质数, 存入vector中if(!vis[i]) primes.push_back(i);     // 注意枚举条件i*primes[j]表示要被筛除的数字(一定不是质数)for(int j = 0; j < primes.size() && i * primes[j] <= n; i++){vis[i * primes[j]] = true;if(i % primes[j] == 0){// 说明往后primes[j]已经不是i*primes[j]的最小质因子了break;}}}
}

时间复杂度分析

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e6 + 5;
bool vis[N]; // vis[i] = true表示被筛掉了
ll prefix[N];
void euler(int n){vector<int> primes;vis[0] = vis[1] = true;for(int i = 2; i <= n; i++){// 如果i没有被筛除, 说明i是质数, 存入vector中if(!vis[i]) primes.push_back(i);     // 注意枚举条件i*primes[j]表示要被筛除的数字(一定不是质数)for(int j = 0; j < primes.size() && i * primes[j] <= n; j++){vis[i * primes[j]] = true;if(i % primes[j] == 0){// 说明往后primes[j]已经不是i*primes[j]的最小质因子了break;}}}
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll n, q; cin >> n >> q;euler(n);for(int i = 1; i <= n; i++) {prefix[i] = prefix[i -1] + (int)(!vis[i] && !vis[i / 2]);}for(int i = 0; i < q; i++){ll l, r; cin >> l >> r;cout << prefix[r] - prefix[l - 1] << "\n";}return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 2e7 + 5;
bool vis[N]; // vis[i] = true表示被筛掉了
ll prefix[N], minp[N];
void euler(int n){vector<int> primes;vis[0] = vis[1] = true;for(int i = 2; i <= n; i++){// 如果i没有被筛除, 说明i是质数, 存入vector中if(!vis[i]) primes.push_back(i), minp[i] = i;     // 注意枚举条件i*primes[j]表示要被筛除的数字(一定不是质数)for(int j = 0; j < primes.size() && i * primes[j] <= n; j++){vis[i * primes[j]] = true;minp[i * primes[j]] = primes[j];if(i % primes[j] == 0){// 说明往后primes[j]已经不是i*primes[j]的最小质因子了break;}}}
}int main(){ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);ll t; cin >> t;euler(2e7);for(int i = 1; i <= 2e7; i++) {prefix[i] = prefix[i -1] + minp[i];}while(t--){ll x; cin >> x;cout << prefix[x] << "\n";}return 0;
}

编程159-161

一分为二

最小质因子之和(简单)

最小质因子之和(困难)

唯一分解定理

唯一分解定理

vector<pair<int, int>> vec;
int main(){int x; cin >> x;// 枚举所有可能的质因子for(int i = 2; i <= x / i; x++){// 如果不能整除直接跳过if(x % i) continue;// 如果可以整除, 那么必然是一个质因子(从小到大的特性决定)// cnt表示当前这个质因子i的指数      int cnt = 0;// 一直除, 知道除干净while(x % i == 0) cnt ++, c /= i;vec.push_back({i, cnt});}if(x > 1) vec.push_back({x, 1});for(const auto i : vec) cout << i.first << " " << i.second << "\n";return 0;
}

约数个数定理

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 105;
ll a[N];void f(int x){// 枚举所有可能的质因子for(int i = 2; i <= x / i; i++){// 如果不能整除直接跳过if(x % i) continue;// 一直除, 知道除干净while(x % i == 0) a[i] ++, x /= i;}if(x > 1) a[x]++;}int main(){int x = 100;for(int i = 1; i <= x; i++){f(i);}ll ans = 1;for(int i = 1; i <= x; i++){ans = ans * (a[i] + 1);}cout << ans << "\n";return 0;
}

约束和定理

编程162-163

没写

快速幂

简介

• 朴素的计算a的b次方的方法, 所需的时间复杂度为O(b)
• 利用倍增的思想, 可以将求幂运算的时间复杂度降低为O(logb)
• 当b为偶数: $a^b=a^{(}b/2)\ast a^{(}b/2)=(a^2{)}^{(}b/2)$
• 当b为奇数: $a^b=a\ast a^{(}b/2)\ast a^{(}b/2)=a\ast (a^2{)}^{(}b/2)$, 注意这里b/2是向下取整
• 于是迭代求解, 直到b为0即可

ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll a[N], n, m, p;ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}void AC(){cin >> n >> m >> p;cout << qmi(n, m) << "\n";
}int main(){ll t; cin >> t;while(t--){AC();}return 0;
}

费马小定理&逆元

费马小定理

逆元

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
const ll p = 1e9 + 7;
ll a[N], n;ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}ll inv(ll x){return qmi(x, p - 2);
}void AC(){cin >> n;cout << inv(n) << "\n";
}int main(){ll t; cin >> t;while(t--){AC();}return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
const ll p = 1e9 + 7;
ll a[N], n, k;ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}ll inv(ll x){return qmi(x, p - 2);
}int main(){cin >> n >> k;if(k == 0){cout << 1 << "\n";for(int i = 2; i <= n; i++){cout << 0 << "\n";}} else if (k & 1){ll ans = inv(n / 2);for(int i = 1; i <= n; i++){if(i & 1) cout << 0 << "\n";else cout << ans << "\n";}} else {ll ans = inv((n + 1) / 2);for(int i = 1; i <= n; i++){if(i & 1) cout << ans << "\n";else cout << 0 << "\n";}}return 0;
}

编程164-166

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll a[N], n, p;ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}ll inv(ll x){return qmi(x, p - 2);
}int main(){cin >> n >> p;cout << inv(n) << "\n";return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll a, b, p;ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}int main(){cin >> a >> b >> p;cout << qmi(a, b) << "\n";return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll a, p;ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}ll inv(ll x){return qmi(x, p - 2);
}int main(){cin >> a >> p;if(a % p == 0){cout << "Inverse doesn't exist\n";} else {cout << inv(a) << "\n";}return 0;
}

欧拉函数&欧拉降幂

欧拉函数

ll phi = n;
for(int i = 2; i <= n; i++){if(n & 1) continue;phi = phi / i * (i - 1);while(n % i == 0) n /= i;
}
if(n > 1) phi = phi / n * (n - 1);

欧拉降幂

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
const ll p = 1e9 + 7;
ll n, m;// 快速幂
ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}int main(){cin >> n >> m;// 欧拉函数ll c = 1e9 + 7, phi = c;for(int i = 2; i <= c / i; i++){if(c % i) continue;phi = phi / i * (i - 1);while(c % i == 0) c /= i;}if(c > 1) phi = phi / c * (c - 1);// 求m的阶乘ll ans = 1;for(int i = 1; i <= m; i++){ans = ans * i % phi; // 为了防止溢出, 每次乘的时候都 % phi}cout << qmi(n, ans) << "\n";return 0;
}

编程167-169

167为乘积幂次那道题

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
const ll p = 1e9 + 7;
ll n;// 欧拉函数
ll euler_phi(ll c){ll phi = c;for(int i = 2; i <= c / i; i++){if(c % i) continue;phi = phi / i * (i - 1);while(c % i == 0) c /= i;}if(c > 1) phi = phi / c * (c - 1);return phi;
}
int main(){cin >> n;cout << euler_phi(n) << "\n";return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll n;// 快速幂
ll qmi(ll a, ll b, ll p){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}// 欧拉函数
ll euler_phi(ll c){ll phi = c;for(int i = 2; i <= c / i; i++){if(c % i) continue;phi = phi / i * (i - 1);while(c % i == 0) c /= i;}if(c > 1) phi = phi / c * (c - 1);return phi;
}int main(){n = 2023;int a = euler_phi(n);for(int i = 2022; i >= 3; i--){n = qmi(i, n, a);}n = qmi(2, n, 2023);cout << n << "\n";return 0;
}

裴蜀定理

简介

证明

推广

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;int main(){ll t; cin >> t;ll a, b; for(int i = 1; i <= t; i++){cin >> a >> b;if(a < 0) a = -a;if(b < 0) b = -b;cout << __gcd(a, b) << "\n";}return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;int main(){ll a, b; cin >> a >> b;cout << a * b - a - b << "\n"; // 裴蜀定理的推广(最大的不能表示的整数)return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll a[N];bool dp[N];int main(){ll n; cin >> n;for(int i = 1; i <= n; i++){cin >> a[i];}sort(a + 1, a + 1 + n);ll d = a[1];for(int i = 2; i <= n; i++){d = __gcd(d, a[i]);}if(d == 1){int m = a[1] * a[2] - a[1] - a[2];dp[0] = true;for(int i = 1; i <= n; i++){for(int j = a[i]; j <= m; j++){dp[j] |= dp[j - a[i]];}}int ans = 0;for(int i = 0; i <= m; i++) ans += (dp[i] == false);cout << ans << "\n";} else {cout << "INF" << "\n";}return 0;
}

编程170-174

171-173是上面例题

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e5  + 5;
ll a[N], GCD;int main(){ll n, m; cin >> n >> m;for(int i = 1; i <= n; i++){cin >> a[i];GCD = __gcd(GCD, a[i]);}cout << m / GCD << "\n";return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 2e5  + 5;
ll a[N], GCD;int main(){ll n, x; cin >> n >> x;for(int i = 1; i <= n; i++){cin >> a[i];GCD = __gcd(GCD, a[i]);}cout << x % GCD << "\n";return 0;
}

组合数学

计数原理

前置基础知识

分类加法

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 2e5  + 5;
const ll p = 1e9 + 7;ll qmi(ll a, ll b){ // 快速幂ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}ll inv(ll x) {return qmi(x, p - 2);} // 逆元ll Cnk(ll n, ll k){ // 排列组合, 这个地方不能使用n的阶乘, 因为会溢出ll res = 1;for(int i = n; i >= n - k + 1; i--) res = res * i % p;for(int i = 1; i <= k; i++) res = res * inv(i) % p;return res;
}ll mo(ll x) {return (x % p + p) % p;} // 取模函数, 防止溢出和防止对分数取模int main(){ll n, a, b; cin >> n >> a >> b;// 性质: Cn0+Cn1+Cn2+...+Cnn = 2^n, Cn0=1, 使用2^n将Cn0和Cna,Cnb减掉即可ll ans = mo(mo(qmi(2ll, n) - 1 - Cnk(n, a)) - Cnk(n, b));cout << ans << "\n";return 0;
}

分步乘法

组合数常用计算方法

DP法求组合数

预处理阶乘法求组合数

差异对比

例题讲解

#include <bits/stdc++.h>
using namespace std;
using ll = long long;vector<vector<ll>> C;int main(){ll n, m, c; cin >> n >> m >> c;C.resize(n + 1);for(int i = 0; i <= n; i++){C[i].resize(m + 1);C[i][0] = 1;}for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % c;}}cout << C[n][m] << "\n";return 0;
}

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll p = 1e9 + 7;
const ll N = 1e7 + 5;ll fac[N], invfac[N];ll qmi(ll a, ll b){ // 快速幂ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}ll inv(ll x) {return qmi(x, p - 2);} // 逆元void init(){int n = 1e7;fac[0] = 1;for(int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % p;invfac[n] = inv(fac[n]);for(int i = n - 1; i >= 0; i--) invfac[i] = (i + 1) * invfac[i + 1] % p;
}ll C(ll n, ll m){ // 排列组合, 这个地方不能使用n的阶乘, 因为会溢出if(n < 0 || m < 0 || n < m) return 0;return fac[n] * invfac[n - m] % p * invfac[m] % p; 
}int main(){ll q; cin >> q;init();while(q--){ll n, m; cin >> n >> m;cout << C(n, m) << "\n";}return 0;
}

排列组合专项

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll p = 1000000007;
const ll N = 1e7 + 5;ll qmi(ll a, ll b){ // 快速幂ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}int main(){ll n, m; cin >> n >> m;cout << (6 * qmi(2, n) + 6 * qmi(2, m) - 24 + p) % p << "\n"; // +p处理负数return 0;
}

二元组计数专项

#include <iostream>
using namespace std;
using ll = long long;
const ll N = 1e5 + 5;
ll a[N], prefix[N], cnt[N]; // cnt[x]表示到当前为止, prefix对k取模后为x的元素个数(包括prefix[0])int main() {ll n, k; cin >> n >> k;for (int i = 1; i <= n; i++) {cin >> a[i];prefix[i] = (prefix[i - 1] + a[i]) % k;}ll ans = 0;cnt[0] = 1;for (ll i = 1; i <= n; i++) {// 维护的是(0, i - 1)的信息ans += cnt[prefix[i]];// 加上i的信息cnt[prefix[i]]++;}cout << ans << "\n";return 0;
}

编程175-181

// 样例过了一半, 还有一半没过#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 5e5 + 5;
const ll p = 1e9 + 7;
ll jc[N], inv_jc[N];ll mo(ll x) {return (x % p + p) % p;} // 取模函数, 防止溢出和防止对分数取模ll qmi(ll a, ll b){ll res = 1;while(b){if(b & 1) res = res * a % p; // b为奇数, 乘一个a到答案里面a = a * a % p, b >>= 1; // 底数平方并对p去模, 指数除以2}return res;
}
ll inv(ll x){return qmi(x, p - 2);
}ll Cnk(ll n, ll k){ // 排列组合, 这个地方不能使用n的阶乘, 因为会溢出ll res = 1;for(ll i = n; i >= n - k + 1; i--) res = res * i % p;for(ll i = 1; i <= k; i++) res = res * inv(i) % p;return res;
}ll A(ll x, ll y){if(y == 0) return 1;else if(x == y) return jc[x];else if(y < 0 || y > x) return 0;else return jc[x] * inv_jc[x - y] % p;
}int main() {ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);jc[0] = 1;for(ll i = 1; i <= N - 5; i++) {jc[i] = jc[i - 1] * i % p;}inv_jc[N - 5] = inv(jc[N - 5]);for(ll i = N - 6; i >= 1; i--){inv_jc[i] = inv_jc[i + 1] * (i + 1) % p;}ll n, m; cin >> n >> m;ll ans = 0;for (ll i = 0; i <= n; i++) {ll f = (i & 1) ? -1 : 1;ans = mo(ans + mo(mo(f * Cnk(n, i)) * A(m - i, n - i)));}cout << ans * A(m, n) % p << "\n";return 0;
}

176-181没写

178, 180上面有