数论基础
扩展欧几里得
hdu2669, 解同余方程 ax+by=1
1 int gcd(int a, int b, int& x, int& y) 2 { 3 if(b == 0){ 4 x = 1; 5 y = 0; 6 return a; 7 } 8 else{ 9 int temp = gcd(b, a % b, x, y); 10 int t = y; 11 y = x - y * (a / b); 12 x = t; 13 return temp; 14 } 15 }
费马小定理 gcd(a,p)=1，那么 a^(p-1) ≡1（mod p）
hdu4828,费马小定理计算卡特兰数
1 #include <iostream> 2 #include <cstdio> 3 #define MAXN 2000001 4 #define MOD 1000000007 5 using namespace std; 6 7 long long f[MAXN + 5]; 8 long long gcd(long long a, long long b, long long& x, long long& y) 9 { 10 if(b == 0){ 11 x = 1; 12 y = 0; 13 return a; 14 } 15 else{ 16 long long temp = gcd(b, a % b, x, y); 17 long long t = y; 18 y = x - y * (a / b); 19 x = t; 20 return temp; 21 } 22 } 23 24 int main() 25 { 26 f[1] = 1; 27 for(long long i = 2; i <= MAXN; i++){ 28 f[i] = (f[i - 1] * (4 * i - 2)) % MOD; 29 gcd(i + 1, MOD, x, y); 30 if(x < 0){ 31 x = MOD - (-x) % MOD; 32 } 33 x = x % MOD; 34 f[i] = (f[i] * x) % MOD; 35 }
中国剩余定理
hdu1573,不互素的中国剩余定理
1 #include <iostream> 2 #include <cstdio> 3 #define M 11 4 using namespace std; 5 6 long long a[M], b[M]; 7 long long x, y; 8 bool flag; 9 long long gcd(long long a, long long b) 10 { 11 if (b == 0){ 12 return a; 13 } 14 return gcd(b, a % b); 15 } 16 17 long long exgcd(long long a, long long b, long long & x, long long & y) 18 { 19 if (b == 0){ 20 x = 1; 21 y = 0; 22 return a; 23 } 24 else{ 25 long long temp = exgcd(b, a % b, x, y); 26 long long t = y; 27 y = x - y * (a / b); 28 x = t; 29 return temp; 30 } 31 } 32 int main() 33 { 34 long long T, d, n, m, t, A, B; 35 scanf("%I64d", &T); 36 for (int cas = 1; cas <= T; cas++){ 37 flag = false; 38 scanf("%I64d%I64d", &n, &m); 39 for (int i = 1; i <= m; i++){ 40 scanf("%d", &a[i]); 41 //t = (t * a[i]) / gcd(t, a[i]); 42 } 43 for (int i = 1; i <= m; i++){ 44 scanf("%I64d", &b[i]); 45 } 46 A = a[1], B = b[1]; 47 for (int i = 2; i <= m; i++){ 48 d = exgcd(A, a[i], x, y); 49 if ((b[i] - B) % d != 0){ 50 flag = true; 51 break; 52 } 53 x = (b[i] - B) / d * x; 54 y = a[i] / d; 55 x = (x % y + y) % y; 56 B = x * A + B; 57 A = (A * a[i]) / d; 58 B = (B % A + A) % A; 59 } 60 if (B > n || flag){ 61 printf("0\n"); 62 } 63 else{ 64 t = 1 + (n - B) / A; 65 if (B == 0){ 66 --t; 67 } 68 printf("%I64d\n", t); 69 } 70 } 71 }
欧拉函数
求1-n与n互质的个数
1 int euler(int x){ 2 int ans = x, m = n; 3 for(int i = 2; i * i <= x; i++){ 4 if(x % i == 0){ 5 ans = ans / i * (i - 1); 6 while(x % i == 0){ 7 x /= i; 8 } 9 } 10 } 11 if(x > 1){ 12 ans = ans / x * (x - 1); 13 } 14 return ans; 15 }
指数循环节
