NOIp 数学 (小学奥数)

Basic knowledge

\[C_n^m=\frac{n!}{m!(n - m)!}
\]

快速幂

// Pure Quickpow
inline int qpow(int n, int m, int mod) {
    ll tot = 1;
    for (ll k = n; m; k = k * k % mod, m >>= 1)
    	if (m & 1) tot = tot * k % mod;
    return tot;
}

/* Matrix Quickpow
 * Au: H15teve
 */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int mod 1000000007

ll n, p;

struct matrix {
    ll m[100][100];
    matrix operator * (matrix &a) {
        matrix b;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) {
                b.m[i][j] = 0;
                for (int k = 0; k < n; k++)
                    b.m[i][j] = (b.m[i][j] + m[i][k] * a.m[k][j]) \% mod;
            }
        return b;
    }
} start;

matrix mpow(matrix a, ll k) {
    if (k == 1) return a;
    a = mpow(a, k / 2);
    if (k \% 2) return (a * a) * start;
    else return a * a;
}

int main() {
    n = readll(), p = readll();
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            start.m[i][j] = readll();

    matrix a = mpow(start, p);

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            writellb(a.m[i][j]);
        writeln();
    }
    return 0;
}

乘法逆元

/* 费马小定理求乘法逆元
 * Au: Menci
 */
inline int qpow(int n, int m, int mod) {
    ll tot = 1;
    for (ll k = n; m; k = k * k % mod, m >>= 1)
    	if (m & 1) tot = tot * k % mod;
    return tot;
}

inline int inv(int x, int mod) {
    return qpow(x, mod - 2);
}

/* 扩展欧几里得求乘法逆元
 * Au: Menci
 */
void exgcd(const int a, const int b, int &g, int &x, int &y) {
    if (!b) g = a, x = 1, y = 0;
    else exgcd(b, a % b, g, y, x), y -= x * (a / b);
}

inline int inv(const int num) {
    int g, x, y;
    exgcd(num, MOD, g, x, y);
    return ((x % MOD) + MOD) % MOD;
}

For more specific explanation, see Link .

组合数

/* Luogu 2822 组合数问题
 * Au: GG
 * C_n^m=\frac{n!}{m!(n - m)!}
 * 预处理 DP O(n^2) + 统计 O(n)
 */
const int N = 2000 + 3, Nx = 2001;
int n, m, t, k, ans, c[N][N], d[N][N];

int main() {
    scanf("%d%d", &t, &k);
    for (int i = 1; i <= Nx; i++) {
        c[i][1] = i % k; c[i][i] = 1;
    }
    for (int i = 2; i <= Nx; i++)
        for (int j = 2; j <= i - 1; j++)
            c[i][j] = (c[i - 1][j] % k + c[i - 1][j - 1] % k) % k;
    for (int i = 1; i <= Nx; i++)
        for (int j = 1; j <= i; j++) {
            if (c[i][j]) d[i][j] = d[i][j - 1];
            else d[i][j] = d[i][j - 1] + 1;
        }
    while (t--) {
        scanf("%d%d", &n, &m);
        ans = 0;
        for (int i = 1; i <= n; i++) {
            if (i > m) ans += d[i][m]; else ans += d[i][i];
        }
        printf("%d\n", ans);
    }

    return 0;
}

同余方程

\[ax \equiv 1 \pmod b
\]

\[ax + by = 1
\]

void exgcd(const int a, const int b, int &g, int &x, int &y) {
    if (!b) g = a, x = 1, y = 0;
    else exgcd(b, a % b, g, y, x), y -= x * (a / b);
}

int main() {
    int a, b, g, x, y;
    scanf("%d%d", &a, &b);
    exgcd(a, b, g, x, y);
    printf("%d\n", (x + b) % b);
    return 0;
}

素数筛法

Eratosthenes 筛法：

/* Sieve of Eratosthenes
 * Au: GG
 */
#include <bits/stdc++.h>
using namespace std;

const int N = 100000002;

int n, prime[N], tot;
bool check[N];

inline void Sieve_of_Eratosthenes() {
    for (register int i = 2; i <= n; i++) {
        if (!check[i]) prime[tot++] = i;
        for (register int j = 0; j < tot; j++) {
            if (i * prime[j] > n) break;
            check[i * prime[j]] = true;
            if (i % prime[j] == 0) break;
        }
    }
}

int main() {
    scanf("%d", &n);
    Sieve_of_Eratosthenes();
    printf("%d\n", tot);
    return 0;
}

GCD

ll gcd(ll x, ll y) {
	if (x == y) return x;
	return !y ? x : gcd(y, x % y);
}

GCD 欧几里得算法

$ a,b $ 为正整数，设集合 \(A=\{xa+yb | x, y\) 是整数 \(\}\)，则 $ A $ 中最小正元素是 $ \gcd(a,b) $

long kgcd(long a, long b) {
	if (a == 0) return b;
	if (b == 0) return a;
	if (!(a & 1) && !(b & 1)) return kgcd(a >> 1, b >> 1) << 1;
	else if (!(b & 1)) return kgcd(a, b >> 1);
	else if (!(a & 1)) return kgcd(a >> 1, b);
	else return kgcd(abs(a - b), min(a, b));
}

LCM

\[\text{lcm} ( a, b ) = a \times b \div \gcd ( a, b )
\]

实际上最好写成 $ a \div \text{lcm} (a,b) \times b $.

long lcm(long a, long b) {
	long c, d, sw;
	c = (a >= b) ? a : b;
	d = (a <= b) ? a : b;
	while (c % d != 0) {
		sw = c % d;
		c = d;
		d = sw;
	}
	return (a / d) * b;
}

求多个数的 \(\textrm{LCM}\)，需要将 \(res\) 初始化为 \(1\)

数的各位之和

int sum(int number) {
	int sum = 0;
	while (number != 0) {
		sum += number % 10;
		number /= 10;
	}
	return sum;
} // 不知道数有几位,但是可以每次都取个位

年份、日期

Reference: Link

/* 判断是否闰年 */
bool isleap(int& year) {
	if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) return true;
	else return false;
}

/* 返回一年的最大天数 */
int maxday(int& year) {
	if (isleap(year)) return 366;
	else return 365;
}

// Days 是指距离某个日期是多少天, 应该均可以的, 只是最终结果可能有所变化的.
string getweek(int& days) {
	return week[days % 7];
}

中国剩余定理

For more specific explanation, see Link .