注：此博客部分模板（也有可能是全部）来源于其它大佬的bolg
1.高精度
1 #include<bits/stdc++.h> 2 #define MAXN 501 3 const int Big_B = 10; const int Big_L = 1; 4 inline int intcmp_ (int a, int b) { if (a > b) return 1; return a < b ? -1 : 0; } 5 struct Int 6 { 7 #define rg register 8 inline int max (int a, int b) { return a > b ? a : b; } 9 inline int min (int a, int b) { return a < b ? a : b; } 10 std :: vector <int> c; Int () {} typedef long long LL; 11 Int (int x) { for (; x > 0; c.push_back (x % Big_B), x /= Big_B); } 12 Int (LL x) { for (; x > 0; c.push_back (x % Big_B), x /= Big_B); } 13 inline void CrZ () { for (; !c.empty () && c.back () == 0; c.pop_back ()); } 14 inline Int &operator += (const Int &rhs){ 15 c.resize (max (c.size (), rhs.c.size ())); rg int i, t = 0, S; 16 for (i = 0, S = rhs.c.size (); i < S; ++ i) 17 c[i] += rhs.c[i] + t, t = c[i] >= Big_B, c[i] -= Big_B & (-t); 18 for (i = rhs.c.size (), S = c.size (); t && i < S; ++ i) 19 c[i] += t, t = c[i] >= Big_B, c[i] -= Big_B & (-t); 20 if (t) c.push_back (t); return *this; 21 } 22 inline Int &operator -= (const Int &rhs){ 23 c.resize (max (c.size (), rhs.c.size ())); rg int i, t = 0, S; 24 for (i = 0, S = rhs.c.size (); i < S; ++ i) 25 c[i] -= rhs.c[i] + t, t = c[i] < 0, c[i] += Big_B & (-t); 26 for (i = rhs.c.size (), S = c.size (); t && i < S; ++ i) 27 c[i] -= t, t = c[i] < 0, c[i] += Big_B & (-t); 28 CrZ (); return *this; 29 } 30 inline Int &operator *= (const Int &rhs){ 31 rg int na = c.size (), i, j, S, ai; 32 c.resize (na + rhs.c.size ()); LL t; 33 for (i = na - 1; i >= 0; -- i){ 34 ai = c[i], t = 0, c[i] = 0; 35 for (j = 0, S = rhs.c.size (); j < S; ++ j){ 36 t += c[i + j] + (LL) ai * rhs.c[j]; 37 c[i + j] = t % Big_B, t /= Big_B; 38 } 39 for (j = rhs.c.size (), S = c.size (); t != 0 && i + j < S; ++ j) 40 t += c[i + j], c[i + j] = t % Big_B, t /= Big_B; 41 assert (t == 0); 42 } 43 CrZ (); return *this; 44 } 45 inline Int &operator /= (const Int &rhs) { return *this = div (rhs); } 46 inline Int &operator %= (const Int &rhs) { return div (rhs), *this; } 47 inline Int &shlb (int l = 1){ 48 if (c.empty ()) return *this; c.resize (c.size () + l);rg int i; 49 for (i = c.size () - 1; i >= l; -- i) c[i] = c[i - l]; 50 for (i = 0; i < l; ++ i) c[i] = 0; 51 return *this; 52 } 53 inline Int &shrb (int l = 1){ 54 for (rg int i = 0; i < c.size () - l; ++ i) c[i] = c[i + l]; 55 c.resize (max (c.size () - l, 0)); return *this; 56 } 57 inline Int div (const Int &rhs){ 58 assert (!rhs.c.empty ()); Int q, r; rg int i; if (rhs > *this) return 0; 59 q.c.resize (c.size () - rhs.c.size () + 1); rg int _l, _r, mid; 60 for (i = c.size () - 1; i > c.size () - rhs.c.size (); -- i) r.shlb (), r += c[i]; 61 for (i = c.size () - rhs.c.size (); i >= 0; -- i){ 62 r.shlb (); r += c[i]; 63 if (r.Comp (rhs) < 0) q.c[i] = 0; 64 else { 65 _l = 0, _r = Big_B; 66 for (; _l != _r; ){ 67 mid = _l + _r >> 1; 68 if ((rhs * mid).Comp (r) <= 0) _l = mid + 1; else _r = mid; 69 } 70 q.c[i] = _l - 1, r -= rhs * q.c[i]; 71 } 72 } 73 q.CrZ (), *this = r; return q; 74 } 75 inline int Comp (const Int &rhs) const { 76 if (c.size () != rhs.c.size ()) return intcmp_ (c.size (), rhs.c.size ()); 77 for (rg int i = c.size () - 1; i >= 0; -- i) 78 if (c[i] != rhs.c[i]) return intcmp_ (c[i], rhs.c[i]); 79 return 0; 80 } 81 friend inline Int operator + (const Int &lhs, const Int &rhs) 82 { Int res = lhs; return res += rhs; } 83 inline friend Int operator - (const Int &lhs, const Int &rhs){ 84 if (lhs < rhs){ 85 putchar ('-'); 86 Int res = rhs; return res -= lhs; 87 } 88 else { Int res = lhs; return res -= rhs; } 89 } 90 friend inline Int operator * (const Int &lhs, const Int &rhs) 91 { Int res = lhs; return res *= rhs; } 92 friend inline Int operator / (const Int &lhs, const Int &rhs) 93 { Int res = lhs; return res.div (rhs); } 94 friend inline Int operator % (const Int &lhs, const Int &rhs) 95 { Int res = lhs; return res.div (rhs), res; } 96 friend inline std :: ostream &operator << (std :: ostream &out, const Int &rhs){ 97 if (rhs.c.size () == 0) out << "0"; 98 else { 99 out << rhs.c.back (); 100 for (rg int i = rhs.c.size () - 2; i >= 0; -- i) 101 out << std :: setfill ('0') << std :: setw (Big_L) << rhs.c[i]; 102 } 103 return out; 104 } 105 friend inline std :: istream &operator >> (std :: istream &in, Int &rhs){ 106 static char s[100000]; 107 in >> s + 1; int Len = strlen (s + 1); 108 int v = 0; LL r = 0, p = 1; 109 for (rg int i = Len; i >= 1; -- i){ 110 ++ v; r = r + (s[i] - '0') * p, p *= 10; 111 if (v == Big_L) rhs.c.push_back (r), r = 0, v = 0, p = 1; 112 } 113 if (v != 0) rhs.c.push_back (r); return in; 114 } 115 friend inline bool operator < (const Int &lhs, const Int &rhs) 116 { return lhs.Comp (rhs) < 0; } 117 friend inline bool operator <= (const Int &lhs, const Int &rhs) 118 { return lhs.Comp (rhs) <= 0; } 119 friend inline bool operator > (const Int &lhs, const Int &rhs) 120 { return lhs.Comp (rhs) > 0; } 121 friend inline bool operator >= (const Int &lhs, const Int &rhs) 122 { return lhs.Comp (rhs) >= 0; } 123 friend inline bool operator == (const Int &lhs, const Int &rhs) 124 { return lhs.Comp (rhs) == 0; } 125 friend inline bool operator != (const Int &lhs, const Int &rhs) 126 { return lhs.Comp (rhs) != 0; } 127 #undef rg 128 }; 129 int Main (){ 130 return 0; 131 } 132 int ZlycerQan = Main (); 133 int main (int argc, char *argv[]) {;}
2.KMP算法
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int next[2333333]; 6 int la, lb, j; 7 char a[2333333], b[2333333]; 8 int main () { 9 cin >> a + 1; 10 cin >> b + 1; 11 la = strlen (a + 1); 12 lb = strlen (b + 1); 13 for (int i = 2; i <= lb; i++) { 14 while (j && b[i] != b[j + 1]) 15 j = next[j]; 16 if (b[j + 1] == b[i])j++; 17 next[i] = j; 18 } 19 j = 0; 20 for (int i = 1; i <= la; i++) { 21 while (j && b[j + 1] != a[i]) 22 j = next[j]; 23 if (b[j + 1] == a[i])j++; 24 if (j == lb) { 25 printf ("%d ", i - lb + 1); 26 j = next[j]; 27 } 28 } 29 for (int i = 1; i <= lb; i++) 30 printf ("%d ", next[i]); 31 return 0; 32 }
求nxt数组是自己匹配自己从2开始
求最大匹配是从1开始遍历被匹配的串
3.加入堆优化的Dijkstra
1 #include <queue> 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 const int N = 1010001; 7 int n, m, s, cnt, head[N], dis[N]; 8 bool vis[N]; 9 struct node{ 10 int next, to, w; 11 }e[N]; 12 int read() { 13 int s = 0, w = 1; 14 char ch = getchar(); 15 while(!isdigit(ch)){if(ch == '-') w = -1;ch = getchar();} 16 while(isdigit(ch)){s = s * 10 + ch - '0';ch = getchar();} 17 return s * w; 18 } 19 void add(int x, int y, int z) { 20 e[++cnt].next = head[x]; 21 e[cnt].to = y; 22 e[cnt].w = z; 23 head[x] = cnt; 24 } 25 struct Node { 26 int u, v; 27 bool operator<(const Node &b) const { 28 return u > b.u; 29 } 30 }; 31 void dijikstra(int s) { 32 priority_queue <Node> q; 33 memset(dis, 0x3f3f3f3f, sizeof(dis)); 34 dis[s] = 0; 35 Node o; 36 o.u = 0; 37 o.v = s; 38 q.push(o); 39 while(!q.empty()) { 40 int u = q.top().v; 41 int d = q.top().u; 42 q.pop(); 43 if(d != dis[u])continue; 44 for(int i = head[u]; i; i = e[i].next) { 45 int v = e[i].to; 46 int w = e[i].w; 47 if(dis[v] > dis[u] + w) { 48 dis[v] = dis[u] + w; 49 Node p; 50 p.u = dis[v], p.v = v; 51 q.push(p); 52 } 53 } 54 } 55 } 56 int main () { 57 n = read(); 58 m = read(); 59 s = read(); 60 while(m--) { 61 int x, y, z; 62 x = read(); 63 y = read(); 64 z = read(); 65 add (x, y, z); 66 } 67 dijikstra(s); 68 for(int i = 1; i <= n; i++) 69 printf("%d ", dis[i]); 70 return 0; 71 } 72 73 堆优化 迪杰
4.未加入堆优化的Dijkstra
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 using namespace std; 5 const int N = 500007; 6 int head[N], cnt, n, m, s, pre[N]; 7 long long dis[N]; 8 bool vis[N]; 9 struct node { 10 int next, to; 11 long long w; 12 }e[N]; 13 void add(int x, int y, long long z) { 14 e[++cnt].next = head[x]; 15 e[cnt].to = y; 16 e[cnt].w = z; 17 head[x] = cnt; 18 } 19 void dijkstra(int s) { 20 for(int i = 1; i <= n; i++) dis[i] = 2147483647; 21 dis[s] = 0; 22 for(int i = 1; i <= n; i++) { 23 int k = 1, maxn = 2147483647; 24 for(int j = 1; j <= n; j++) 25 if(!vis[j] && dis[j] <= maxn) 26 k = j, maxn = dis[j]; 27 vis[k] = 1; 28 for(int j = head[k]; j; j = e[j].next) 29 if(dis[e[j].to] > dis[k] + e[j].w) 30 dis[e[j].to] = dis[k] + e[j].w/*, pre[e[j].to] = k*/; 31 } 32 } 33 int main () { 34 scanf("%d%d%d", &n, &m, &s); 35 for(int i = 1; i <= m; i++) { 36 int x, y; 37 long long z; 38 scanf("%d%d%lld", &x, &y, &z); 39 add(x, y, z); 40 } 41 dijkstra(s); 42 for(int i = 1; i <= n; i++) 43 printf("%lld ", dis[i]); 44 /*while(pre[n]) { 45 cout << pre[n] << " "; 46 n = pre[n]; 47 }*/ 48 return 0; 49 }
注释掉的是记录路径的方法
5.SPFA
1 #include <iostream> 2 #include <cstdio> 3 #include <queue> 4 #define N 500005 5 #define inf 2147483647 6 using namespace std; 7 int n, m, s, cnt; 8 int dis[N], vis[N], head[N]; 9 struct node { 10 int next, to, w; 11 }tr[N]; 12 void add (int x, int y, int z) { 13 tr[++cnt].to = y; 14 tr[cnt].next = head[x]; 15 tr[cnt].w = z; 16 head[x] = cnt; 17 } 18 void spfa () { 19 queue<int> q; 20 for (int i = 1; i <= n; i++) 21 dis[i] = inf; 22 vis[s] = 1; 23 q.push(s); 24 dis[s] = 0; 25 while (!q.empty()) { 26 int he = q.front(); 27 q.pop(); 28 vis[he] = 0; 29 for (int i = head[he]; i ;i = tr[i].next) { 30 if (dis[tr[i].to] > dis[he] + tr[i].w) { 31 dis[tr[i].to] = dis[he] + tr[i].w; 32 if (!vis[tr[i].to]) { 33 vis[tr[i].to] = 1; 34 q.push(tr[i].to); 35 } 36 } 37 } 38 } 39 } 40 int main () { 41 scanf ("%d%d%d", &n, &m, &s); 42 for (int i = 1; i <= m; i++) { 43 int a, b, c; 44 scanf ("%d%d%d", &a, &b, &c); 45 add (a, b, c); 46 } 47 spfa (); 48 for (int i = 1; i <= n; i++) 49 if (s == i) printf ("0 "); 50 else printf ("%d ", dis[i]); 51 return 0; 52 } 53 54 spfa
6.Tire树
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 char ch[800000]; 6 int n, m, tr[8000000][26], s, root, tot, f[8000000]; 7 int main () { 8 scanf ("%d", &n); 9 for (int i = 1; i <= n; i++) { 10 root = 0; 11 cin >> ch; 12 int len = strlen (ch); 13 for (int j = 0; j < len; j++) { 14 int nu = ch[j] - 'a'; 15 if (!tr[root][nu]) tr[root][nu] = ++tot; 16 root = tr[root][nu]; 17 } 18 f[root] = 1; 19 } 20 scanf ("%d", &m); 21 for (int i = 1; i <= m; i++) { 22 root = 0; 23 cin >> ch; 24 int len = strlen (ch); 25 for (int j = 0; j < len; j++) { 26 int nu = ch[j] - 'a'; 27 if (!tr[root][nu]) { 28 printf ("WRONG "); 29 break; 30 } 31 root = tr[root][nu]; 32 if (j == len - 1 && f[root] == 2) printf ("REPEAT "); 33 if (j == len - 1 && f[root] == 1) f[root] = 2,printf ("OK "); 34 } 35 } 36 return 0; 37 }
7.树状数组
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int n, m, tr[1000000]; 5 int lowbit (int x) { 6 return x & (-x); 7 } 8 void add (int x, int k) { 9 while (x <= n) { 10 tr[x] += k; 11 x += lowbit (x); 12 } 13 } 14 int sum (int x) { 15 int ans = 0; 16 while (x !=0) { 17 ans += tr[x]; 18 x -= lowbit (x); 19 } 20 return ans; 21 } 22 int main () { 23 scanf ("%d%d", &n, &m); 24 for (int i = 1; i <= n; i++) { 25 int a; 26 scanf ("%d", &a); 27 add (i, a); 28 } 29 for (int i = 1; i <= m; i++) { 30 int a, b, c; 31 scanf ("%d%d%d", &a, &b, &c); 32 if (a == 1) add (b, c); 33 if (a == 2) cout << sum(c) - sum (b - 1) << endl; 34 } 35 return 0; 36 }
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 int n, m, now, x, c[500001]; 5 int lowbit (int x) { 6 return x & -x; 7 } 8 void build (int x, int y) { 9 while (x <= n) { 10 c[x] += y; 11 x += lowbit (x); 12 } 13 } 14 int sum (int x) { 15 int ans = 0; 16 while (x) { 17 ans += c[x]; 18 x -= lowbit (x); 19 } 20 return ans; 21 } 22 int main () { 23 scanf ("%d%d", &n, &m); 24 for (int i = 1; i <= n; i++) { 25 scanf ("%d", &x); 26 build (i, x - now); 27 now = x; 28 } 29 while (m--) { 30 scanf ("%d", &x); 31 if (x == 1) { 32 int l, r, w; 33 scanf ("%d%d%d", &l, &r, &w); 34 build (l, w); 35 build (r + 1, -w); 36 } 37 else if (x == 2) { 38 scanf ("%d", &x); 39 printf ("%d ", sum (x)); 40 } 41 } 42 return 0; 43 }
8.裴蜀定理
1 #include <cstdio> 2 #include <iostream> 3 using namespace std; 4 int n, ans, x, y; 5 int read() { 6 int s = 0, w = 1; 7 char ch = getchar(); 8 while(!isdigit(ch)) {if(ch == '-') w = -1; ch = getchar();} 9 while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = getchar();} 10 return s * w; 11 } 12 int gcd(int x, int y) { 13 // if(x < 0) x = -x; 14 // if(y < 0) y = -y; 15 return y == 0 ? x : gcd(y, x % y); 16 } 17 int main() { 18 n = read(); 19 x = read(), y = read(); 20 if(x < 0) x = -x; 21 if(y < 0) y = -y; 22 ans = gcd(x, y); 23 for(int i = 1; i <= n - 2; i++) { 24 y = read(); 25 if(y < 0) y = -y; 26 ans = gcd(ans, y); 27 // x = ans; 28 } 29 cout << ans << endl; 30 return 0; 31 }
9.gcd
1 int gcd(int x, int y) { 2 return y == 0 ? x : gcd(y, x % y); 3 }
10.lcm（最小公倍数）
众所周知lcm(x, y) = x * y / gcd(x, y)
11.exgcd
1 void exgcd(int a, int b, int &x, int &y) { 2 if(!b) {x = 1, y = 0;return;} 3 int d = exgcd(b, a % b, y, x); 4 y -= a / b * x; 5 return d; 6 }
12.欧拉筛
#include<iostream> #include<cstdio> using namespace std; int n, p[1000], vis[1000],num; int oulashai(int x) { for(int i = 2; i <= n; i++) { if(!vis[i])p[++num] = i; for(int j = 1; j <= num; j++) { if(i * p[j] > n) break; vis[i * p[j]] = 1; if(i % p[j] == 0) break; } } } int main() { scanf("%d",&n); oulashai(n); for(int i = 1; i <= num; i++) printf("%d ",p[i]); }
就先整理到这里趴你们的博主太菜了