huanghai/python
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

'commit'

Browse Source
This commit is contained in:
HuangHai
2025-08-30 18:35:01 +08:00
commit 1f397eca87
6258 changed files with 1487241 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

78
TangDou/AcWing/Math/ZhuHeJiShu/1257.cpp Normal file
View File

@@ -0,0 +1,78 @@
#include <bits/stdc++.h>
using namespace std;
const int N = 10010; // 5000*2+10
int primes[N], cnt;
bool st[N];
int a[N], b[N];
void get_primes(int n) {
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 get(int n, int p) { // n!中p的次数
int s = 0;
while (n) n /= p, s += n;
return s;
}
void mul(int a[], int b, int &len) {
int t = 0;
for (int i = 1; i <= len; i++) {
t += a[i] * b;
a[i] = t % 10;
t /= 10;
}
while (t) {
a[++len] = t % 10;
t /= 10;
}
//去前导0
while (len > 1 && !a[len]) len--;
}
int C(int a, int b, int c[]) {
int len = 1;
c[1] = 1;
for (int i = 0; i < cnt; i++) {
int p = primes[i];
int s = get(a, p) - get(b, p) - get(a - b, p);
while (s--) mul(c, p, len);
}
return len;
}
void sub(int a[], int b[], int &len) {
int t = 0;
for (int i = 1; i <= len; i++) {
t = a[i] - b[i] - t;
a[i] = (t + 10) % 10;
t < 0 ? t = 1 : t = 0;
}
while (len > 1 && !a[len]) len--;
}
int main() {
//加快读入
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
get_primes(N - 10);
int n;
cin >> n;
int a1 = C(n + n, n, a);
int b1 = C(n + n, n - 1, b); // bl下面没有用到，原因是两数相减，我们知道a>b,按着a的长度来就行了
sub(a, b, a1);
for (int i = a1; i >= 1; i--) printf("%d", a[i]);
return 0;
}

78
TangDou/AcWing/Math/ZhuHeJiShu/1317.cpp Normal file
View File

@@ -0,0 +1,78 @@
#include <bits/stdc++.h>
using namespace std;
const int N = 1010; // n+n=2n=> 500*2+10
int primes[N], cnt;
bool st[N];
int a[N], b[N];
void get_primes(int n) {
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 get(int n, int p) { // n!中p的次数
int s = 0;
while (n) n /= p, s += n;
return s;
}
void mul(int a[], int b, int &len) {
int t = 0;
for (int i = 1; i <= len; i++) {
t += a[i] * b;
a[i] = t % 10;
t /= 10;
}
while (t) {
a[++len] = t % 10;
t /= 10;
}
//去前导0
while (len > 1 && !a[len]) len--;
}
int C(int a, int b, int c[]) {
int len = 1;
c[1] = 1;
for (int i = 0; i < cnt; i++) {
int p = primes[i];
int s = get(a, p) - get(b, p) - get(a - b, p);
while (s--) mul(c, p, len);
}
return len;
}
void sub(int a[], int b[], int &len) {
int t = 0;
for (int i = 1; i <= len; i++) {
t = a[i] - b[i] - t;
a[i] = (t + 10) % 10;
t < 0 ? t = 1 : t = 0;
}
while (len > 1 && !a[len]) len--;
}
int main() {
//加快读入
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
get_primes(2 * n);
int a1 = C(n + n, n, a);
int b1 = C(n + n, n - 1, b); // bl下面没有用到，原因是两数相减，我们知道a>b,按着a的长度来就行了
sub(a, b, a1);
for (int i = a1; i >= 1; i--) printf("%d", a[i]);
return 0;
}

98
TangDou/AcWing/Math/ZhuHeJiShu/1317.md Normal file
View File

@@ -0,0 +1,98 @@
AcWing 1317. 树屋阶梯
[题目传送门](https://www.acwing.com/problem/content/description/1319/)
### 一、题目分析
考虑以阶梯左下角那个点为第一个钢材的左下角，那么第一个钢材摆放情况便如下图（以 $n = 5$ 为例）
<center><img src='https://img-blog.csdnimg.cn/8e09382d5838423fa239a81e00ab40f2.png?x-oss-process=image/watermark,type_ZHJvaWRzYW5zZmFsbGJhY2s,shadow_50,text_Q1NETiBAeWV6enou,size_20,color_FFFFFF,t_70,g_se,x_16'></center>
对每种情况分别讨论，那么问题都被分成了两个子问题，设$f[n]$表示摆放高度为$n$的台阶的方法数，那么：
$$\large f[5]=f[4]*f[0]+f[3]*f[1]+f[2]*f[2]+f[1]*f[3]+f[0]*f[4]=\sum_{k=1}^5f(5-k)*f(k-1)$$
泛化一下，就是：
$$\large f[n]=f[n-1]*f[0]+f[n-2]*f[1]+f[n-3]*f[2]+...+f[1]*f[3]+f[0]*f[n-1]=\sum_{k=1}^nf(n-k)*f(k-1)$$
显然，这就是**卡特兰数**了
但是这题很烦，还要用高精乘:
```c++
#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
int primes[N], cnt;
bool st[N];
int a[N], b[N];
void get_primes(int n) {
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 get(int n, int p) { // n!中p的次数
int s = 0;
while (n) n /= p, s += n;
return s;
}
void mul(int a[], int b, int &len) {
int t = 0;
for (int i = 1; i <= len; i++) {
t += a[i] * b;
a[i] = t % 10;
t /= 10;
}
while (t) {
a[++len] = t % 10;
t /= 10;
}
//去前导0
while (len > 1 && !a[len]) len--;
}
int C(int a, int b, int c[]) {
int len = 1;
c[1] = 1;
for (int i = 0; i < cnt; i++) {
int p = primes[i];
int s = get(a, p) - get(b, p) - get(a - b, p);
while (s--) mul(c, p, len);
}
return len;
}
void sub(int a[], int b[], int &len) {
int t = 0;
for (int i = 1; i <= len; i++) {
t = a[i] - b[i] - t;
a[i] = (t + 10) % 10;
t < 0 ? t = 1 : t = 0;
}
while (len > 1 && !a[len]) len--;
}
int main() {
//加快读入
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
get_primes(N - 10);
int n;
cin >> n;
int a1 = C(n + n, n, a);
int b1 = C(n + n, n - 1, b); // bl下面没有用到，原因是两数相减，我们知道a>b,按着a的长度来就行了
sub(a, b, a1);
for (int i = a1; i >= 1; i--) printf("%d", a[i]);
return 0;
}
```

65
TangDou/AcWing/Math/ZhuHeJiShu/CatalanNumber.cpp Normal file
View File

@@ -0,0 +1,65 @@
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int mod = 1e9 + 7;
const int N = 110;
int f[N];
int qmi(int a, int k, int p) {
int res = 1;
while (k) {
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
int C(int a, int b, int p) {
if (a < b) return 0;
int down = 1, up = 1;
for (int i = a, j = 1; j <= b; i--, j++) {
up = (LL)up * i % p;
down = (LL)down * j % p;
}
return (LL)up * qmi(down, p - 2, p) % p;
}
//卡特兰数的前几项为：1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786,208012
int main() {
//目标：求出卡特兰数的前12位
//公式法计算I
//方法1:f(n)=C(n,2n)-C(n-1,2n)
for (int i = 1; i <= 12; i++)
cout << C(2 * i, i, mod) - C(2 * i, i - 1, mod) << " ";
cout << endl;
//公式法计算II
//方法2：C(n,2n)/(n+1)
for (int i = 1; i <= 12; i++)
cout << C(2 * i, i, mod) / (i + 1) << " ";
cout << endl;
//递推法计算I
//方法3：f(n)=(4n-2)/(n+1)*f(n-1)
f[0] = 1;
for (int i = 1; i <= 12; i++) {
f[i] = (LL)f[i - 1] * (4 * i - 2) / (i + 1); //注意：这里要先乘再除，否则可能丢失精度
cout << f[i] << " ";
}
cout << endl;
//递推法计算II
//方法4： f(n)=\sum_{i=0}^{n-1}f(i)*f(n-i+1)
//要是每次都这么推，不是效率太差了吗？
memset(f, 0, sizeof f);
f[0] = 1;
for (int i = 1; i <= 12; i++) {
for (int j = 1; j <= i; j++)
f[i] += f[j - 1] * f[i - j];
cout << f[i] << " ";
}
return 0;
}

18
TangDou/AcWing/Math/ZhuHeJiShu/CatalanNumberPaiDui.cpp Normal file
View File

@@ -0,0 +1,18 @@
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL C(int n, int m) {
LL a = 1, b = 1;
for (int i = 0; i < m; i++) {
a *= (n - i);
b *= (m - i);
}
return a / b;
}
int main() {
int n;
cin >> n; // n过大会溢出，需要结合高精度
printf("%lld\n", C(2 * n, n) / (n + 1));
return 0;
}

16
TangDou/AcWing/Math/ZhuHeJiShu/P1044.cpp Normal file
View File

@@ -0,0 +1,16 @@
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 20;
int f[N];
int main() {
int n;
cin >> n;
f[0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
f[i] += f[j - 1] * f[i - j];
cout << f[n] << endl;
return 0;
}

20
TangDou/AcWing/Math/ZhuHeJiShu/P1645.cpp Normal file
View File

@@ -0,0 +1,20 @@
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1010, MOD = 1e9 + 7;
int n;
int f[N];
int main() {
cin >> n;
f[0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
f[i] = (f[i] + (LL)f[j - 1] * f[i - j]) % MOD;
cout << f[n] << endl;
return 0;
}

16
TangDou/AcWing/Math/ZhuHeJiShu/P1754.cpp Normal file
View File

@@ -0,0 +1,16 @@
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 30;
LL f[N];
int main() {
int n;
cin >> n;
f[0] = 1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
f[i] += f[j - 1] * f[i - j];
cout << f[n] << endl;
return 0;
}

33
TangDou/AcWing/Math/ZhuHeJiShu/Test.cpp Normal file
View File

@@ -0,0 +1,33 @@
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long LL;
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
int fact[N];
int infact[N];
int qmi(int a, int k, int p) {
int res = 1;
while (k) {
if (k & 1) res = (LL)res * a % p;
a = (LL)a * a % p;
k >>= 1;
}
return res;
}
int main() {
fact[0] = infact[0] = 1;
for (int i = 1; i < N; i++) {
fact[i] = (LL)fact[i - 1] * i % mod;
infact[i] = (LL)infact[i - 1] * qmi(i, mod - 2, mod) % mod;
}
int n;
scanf("%d", &n);
while (n--) {
int a, b;
scanf("%d %d", &a, &b);
printf("%d\n", (LL)fact[a] * infact[b] % mod * infact[a - b] % mod);
}
return 0;
}