我的模板（持续更新ing）[早就不更新了]

头文件

短的：

 1 #include<bits/stdc++.h>
 2 #define cl(a,b) memset(a,b,sizeof(a))
 3 #define debug(a) cerr<<#a<<"=="<<a<<endl
 4 using namespace std;
 5 typedef long long ll;
 6 typedef pair<int,int> pii;
 7 
 8 const int maxn=1e5+10;
 9 
10 int main()
11 {
12 
13     return 0;
14 }/*
15 
16 
17 
18 */
View Code

 

长的：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<string>
 7 #include<ctime>
 8 #include<map>
 9 #include<set>
10 #include<vector>
11 #include<queue>
12 #include<cstdlib>
13 #include<cassert>
14 #include<sstream>
15 #include<stack>
16 #include<list>
17 #include<bitset>
18 #define cl(a,b) memset(a,b,sizeof(a))
19 #define debug(x) cerr<<#x<<"=="<<(x)<<endl
20 using namespace std;
21 typedef long long ll;
22 typedef long double ldb;
23 typedef pair<int,int> pii;
24 
25 const int inf=0x3f3f3f3f;
26 const int maxn=1e9+10;
27 const int mod=1e7+7;
28 const double eps=1e-8;
29 const double pi=acos(-1);
30 
31 int dx[8]= {0,0,1,-1,1,-1,1,-1};
32 int dy[8]= {1,-1,0,0,-1,1,1,-1};
33 
34 ll gcd(ll a,ll b){return a?gcd(b%a,a):b;}
35 ll powmod(ll a,ll x,ll mod){ll t=1;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
36 //---------------------------------------ヽ(^。^)丿
37 int main()
38 {
39 
40     return 0;
41 }
42 /*
43 
44 
45 
46 */
View Code

 

超神读入挂

 1 #define JUDGE
 2 
 3 #ifdef DEBUG
 4 inline bool read(int &ret)
 5 {
 6     if(scanf("%d",&ret)==EOF) return false;
 7     return true;
 8 }
 9 #endif  DEBUG
10 
11 #ifdef JUDGE
12 namespace fastIO {
13     #define BUF_SIZE 100000
14     //fread -> read
15     bool IOerror = 0;
16     inline char nc() {
17         static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
18         if(p1 == pend) {
19             p1 = buf;
20             pend = buf + fread(buf, 1, BUF_SIZE, stdin);
21             if(pend == p1) {
22                 IOerror = 1;
23                 return -1;
24             }
25         }
26         return *p1++;
27     }
28     inline bool blank(char ch) {
29         return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
30     }
31     inline bool read(int &x) {
32         char ch;
33         while(blank(ch = nc()));
34         if(IOerror)
35             return false;
36         for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
37         return true;
38     }
39     #undef BUF_SIZE
40 };
41 using namespace fastIO;
42 #endif JUDGE
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矩阵快速幂

 1 struct matrix
 2 {
 3     ll a[maxn][maxn];
 4     int row,col;
 5     matrix():row(maxn),col(maxn){cl(a,0);}
 6     matrix(int x,int y):row(x),col(y){cl(a,0);}
 7     void show()
 8     {
 9         for(int i=0; i<row; i++)
10         {
11             for(int j=0; j<col; j++)
12             {
13                 printf("%d%c",a[i][j],j==col-1?'\n':' ');
14             }
15         }
16     }
17     inline ll* operator [] (int x)
18     {
19         return a[x];
20     }
21     inline matrix operator * (matrix x)
22     {
23         matrix tmp(row,col);
24         for (int i=0; i<row; i++)
25             for (int j=0; j<col; j++)
26                 for (int k=0; k<row; k++)
27                     tmp[i][j]=(tmp[i][j]+a[i][k]*x[k][j])%mod;
28         return tmp;
29     }
30     inline void operator *= (matrix x)
31     {
32         *this = *this * x;
33     }
34     matrix operator ^ (ll x)
35     {
36         matrix result(row,col),now=*this;
37         for(int i=0; i<row; i++)
38         {
39             result[i][i]=1;
40         }
41         while(x)
42         {
43             if(x%2) result*=now;
44             now*=now;
45             x/=2;
46         }
47         return result;
48     }
49 };
View Code

 

降维线段树（区间更新+单点更新）

  1 #define ls getid(l,l+r>>1)
  2 #define rs getid((l+r>>1)+1,r)
  3 #define lson l,m,ls
  4 #define rson m+1,r,rs
  5 
  6 typedef long long ll;
  7 
  8 const int maxn = 1e5+10;
  9 
 10 inline int getid(int x,int y)
 11 {
 12     return x+y|y!=x;
 13 }
 14 
 15 ll rm[maxn<<1],col[maxn<<1];
 16 
 17 void pushup(int l,int r,int rt)
 18 {
 19     rm[rt]=rm[ls]+rm[rs];
 20 }
 21 
 22 void pushdown(int l,int r,int rt,int k)
 23 {
 24     if (col[rt])
 25     {
 26         col[ls] += col[rt];
 27         col[rs] += col[rt];
 28         rm[ls] += col[rt]*(k-(k>>1));
 29         rm[rs] += col[rt]*(k>>1);
 30         col[rt] = 0;
 31     }
 32 }
 33 
 34 void build(int l,int r,int rt)
 35 {
 36     if(l == r)
 37     {
 38         scanf("%I64d",&rm[rt]);
 39         return ;
 40     }
 41     int m=l+r>>1;
 42     build(lson);
 43     build(rson);
 44     pushup(l,r,rt);
 45 }
 46 
 47 void update(int L,int R,int c,int l,int r,int rt)
 48 {//区间更新
 49     if(L<=l && r<=R)
 50     {
 51         col[rt]+=c;
 52         rm[rt]+=c*(r-l+1);
 53         return ;
 54     }
 55     pushdown(l,r,rt,r-l+1);
 56     int m=l+r>>1;
 57     if( L <= m ) update(L,R,c,lson);
 58     if (R > m) update(L,R,c,rson);
 59     pushup(l,r,rt);
 60 }
 61 
 62 void update(int p,int c,int l,int r,int rt)
 63 {//单点更新
 64     if(l==r)
 65     {
 66         rm[rt]+=c;
 67         return ;
 68     }
 69     int m=l+r>>1;
 70     if( p <= m ) update(p,c,lson);
 71     else update(p,c,rson);
 72     pushup(l,r,rt);
 73 }
 74 
 75 ll query(int L,int R,int l,int r,int rt)
 76 {
 77     if( L<=l && r<=R )
 78     {
 79         return rm[rt];
 80     }
 81     pushdown(l,r,rt, r-l+1);
 82     int m=l+r>>1;
 83     ll ret = 0;
 84     if( L<=m ) ret += query(L,R,lson);
 85     if( m<R ) ret += query(L,R,rson);
 86     return ret;
 87 }
 88 
 89 int main()
 90 {
 91     int n,m;
 92     scanf("%d%d",&n,&m);
 93     char str[2];
 94     build(1,n,1);
 95     while(m--)
 96     {
 97         int l,r,x;
 98         scanf("%s",str);
 99         if(str[0]=='Q')
100         {
101             scanf("%d%d",&l,&r);
102             printf("%I64d\n",query(l,r,1,n,1));
103         }
104         else
105         {
106             scanf("%d %d %d", &l, &r, &x);
107             update(l, r, x, 1, n, 1);
108         }
109     }
110     return 0;
111 }
View Code

 

字典树求异或值

 1 typedef long long ll;
 2 
 3 const int maxn=1e5+10;
 4 
 5 struct t
 6 {
 7     ll tree[32*maxn][3];//字典树的树部分
 8     ll val[32*maxn];//字典树的终止节点
 9     ll cnt;//字典树的大小
10     void init() //初始化字典树
11     {
12         cl(tree,0);
13         cl(val,0);
14         cnt=1;
15     }
16     void add(ll a)
17     {
18         int x=0,now;
19         for(int i=32; i>=0; i--)
20         {
21             now=((a>>i)&1); //当前位数
22             if(tree[x][now]==0) //没有的这个节点的话 加入这个节点
23             {
24                 cl(tree[cnt],0);
25                 val[cnt]=0;
26                 tree[x][now]=cnt++;
27             }
28             x=tree[x][now];
29         }
30         val[x]=a; //结尾标记
31     }
32     ll Search(ll a) //查找字典树中和a异或起来最大的值
33     {
34         int x=0,now;
35         for(int i=32; i>=0; i--)
36         {
37             now=!((a>>i)&1);
38             if(tree[x][now]) //如果和当前位相反的这个节点存在
39             {
40                 x=tree[x][now]; //那么继续找这个节点下的子树
41             }
42             else
43             {
44                 x=tree[x][!now]; //不存在就找自己这个节点下的子树
45             }
46         }
47         return val[x]; //结尾标记就是这个数字
48     }
49 } Trie;
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马拉车算法求最长回文串

 1 string longestPalindrome(string ss)
 2 {
 3     string s="#";
 4     for(auto i:ss)
 5     {
 6         s+=i;
 7         s+="#";
 8     }
 9     int c=0,r=0;
10     vector<int>p(s.size(),0);
11     for(int i=0;i<s.size();i++)
12     {
13         if(r>i) p[i]=min(r-i,p[2*c-i]);
14         while((i-1-p[i])>=0
15             &&(i+1+p[i])<s.size()
16             &&s[i+1+p[i]]==s[i-1-p[i]]) 
17                 p[i]++;
18         if(i+p[i]>r)
19         {
20             r=i+p[i];
21             c=i;
22         }
23     }
24     int ans=0,mid=0;
25     for(int i=0;i<s.size();i++)
26     {
27         if(p[i]>ans)
28         {
29             ans=p[i];
30             mid=i;
31         }
32     }
33     return ss.substr((mid-ans)/2,ans);
34 }
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lucas定理

 1 ll fac[maxn];
 2 
 3 ll quick_pow(ll a,ll n)
 4 {
 5     ll ans=1;
 6     while(n)
 7     {
 8         if(n%2) ans=1ll*ans*a%mod;
 9         a=1ll*a*a%mod;
10         n>>=1;
11     }
12     return ans;
13 }
14 
15 ll inv(ll a)
16 {
17     return quick_pow(a,mod-2);
18 }
19 
20 void init()
21 {
22     fac[0]=1;
23     for(int i=1;i<maxn;i++)
24     {
25         fac[i]=1ll*fac[i-1]*i%mod;
26     }
27 }
28 
29 ll C(ll a,ll b)
30 {
31     ll ans=1;
32     if(a<b || b<0) return 0;
33     while(a&&b)
34     {
35         ll aa=a%mod,bb=b%mod;
36         if(aa<bb) return 0;;
37         ans = 1ll*ans*fac[aa]%mod * inv(1ll*fac[bb]*fac[aa-bb]%mod)%mod;
38         a/=mod,b/=mod;
39     }
40     return ans;
41 }
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O(n)预处理Cm n

 1 const int maxn=1e6+10;
 2 const int mod=1e9+7;
 3 ll fac[maxn],f[maxn],inv[maxn];
 4 
 5 void init()
 6 {
 7     fac[0]=fac[1]=f[0]=f[1]=inv[0]=inv[1]=1;
 8     for(int i=2;i<maxn;i++)
 9     {
10         fac[i]=fac[i-1]*i%mod;   // n！
11         ll t=mod/i,k=mod%i;
12         f[i]=(mod-t)*f[k]%mod;  // n的逆元
13         inv[i]=inv[i-1]*f[i]%mod; // n！的逆元
14     }
15 }
16 
17 ll C(ll n,ll m)
18 {
19     if(n<m) return 0;
20     return fac[n]*inv[m]%mod*inv[n-m]%mod;
21 }
View Code

 

自适应simpson积分

 1 double simpson(double a,double b)
 2 {
 3     double c=a+(b-a)/2;
 4     return (f(a)+4*f(c)+f(b))*(b-a)/6;
 5 }
 6 
 7 double asr(double a,double b,double eps,double A)
 8 {
 9     double c=a+(b-a)/2;
10     double L=simpson(a,c);
11     double R=simpson(c,b);
12     if(fabs(L+R-A)<=15*eps) return L+R+(L+R-A)/15.0;
13     return asr(a,c,eps/2,L)+asr(c,b,eps/2,R);
14 }
15 
16 double asr(double a,double b,double eps)
17 {
18     return asr(a,b,eps,simpson(a,b));
19 }
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日期公式

 1 int zeller(int y,int m,int d)
 2 {//计算星期几
 3     if(m<=2) y--,m+=12;
 4     int c=y/100;y%=100;
 5     int w=((c>>2)-(c<<1)+y+(y>>2)+(13*(m+1)/5)+d-1)%7;
 6     if(w<0) w+=7; return (w);
 7 }
 8 
 9 int getid(int y,int m,int d)
10 {//计算每个日期的id 可以计算两个日期之间的天数
11     if(m<3) {y--;m+=12;}
12     return 365*y+y/4-y/100+y/400+(153*m+2)/5+d;
13 }
View Code

 

费用流（uva11248）

  1 // UVa11248 Frequency Hopping：使用Dinic算法
  2 // Rujia Liu
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<queue>
  6 #include<vector>
  7 #include<algorithm>
  8 using namespace std;
  9 
 10 const int maxn = 100 + 10;
 11 const int INF = 1000000000;
 12 
 13 struct Edge
 14 {
 15     int from, to, cap, flow;
 16     bool operator < (const Edge& b)
 17     {
 18         return from < b.from || (from == b.from && to < b.to);
 19     }
 20 };
 21 
 22 struct Dinic
 23 {
 24     int n, m, s, t;         //点的编号从0开始
 25     vector<Edge> edges;    // 边数的两倍
 26     vector<int> G[maxn];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
 27     bool vis[maxn];         // BFS使用
 28     int d[maxn];           // 从起点到i的距离
 29     int cur[maxn];        // 当前弧指针
 30 
 31     void ClearAll(int n)
 32     {
 33         for(int i = 0; i < n; i++) G[i].clear();
 34         edges.clear();
 35     }
 36 
 37     void ClearFlow()
 38     {
 39         for(int i = 0; i < edges.size(); i++) edges[i].flow = 0;
 40     }
 41 
 42     void AddEdge(int from, int to, int cap)
 43     {
 44         edges.push_back((Edge){from, to, cap, 0});
 45         edges.push_back((Edge){to, from, 0, 0});
 46         m = edges.size();
 47         G[from].push_back(m-2);
 48         G[to].push_back(m-1);
 49     }
 50 
 51     bool BFS()
 52     {
 53         memset(vis, 0, sizeof(vis));
 54         queue<int> Q;
 55         Q.push(s);
 56         vis[s] = 1;
 57         d[s] = 0;
 58         while(!Q.empty())
 59         {
 60             int x = Q.front();
 61             Q.pop();
 62             for(int i = 0; i < G[x].size(); i++)
 63             {
 64                 Edge& e = edges[G[x][i]];
 65                 if(!vis[e.to] && e.cap > e.flow)
 66                 {
 67                     vis[e.to] = 1;
 68                     d[e.to] = d[x] + 1;
 69                     Q.push(e.to);
 70                 }
 71             }
 72         }
 73         return vis[t];
 74     }
 75 
 76     int DFS(int x, int a)
 77     {
 78         if(x == t || a == 0) return a;
 79         int flow = 0, f;
 80         for(int& i = cur[x]; i < G[x].size(); i++)
 81         {
 82             Edge& e = edges[G[x][i]];
 83             if(d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0)
 84             {
 85                 e.flow += f;
 86                 edges[G[x][i]^1].flow -= f;
 87                 flow += f;
 88                 a -= f;
 89                 if(a == 0) break;
 90             }
 91         }
 92         return flow;
 93     }
 94 
 95     int Maxflow(int s, int t)
 96     {
 97         this->s = s;
 98         this->t = t;
 99         int flow = 0;
100         while(BFS())
101         {
102             memset(cur, 0, sizeof(cur));
103             flow += DFS(s, INF);
104         }
105         return flow;
106     }
107 
108     vector<int> Mincut()   // call this after maxflow
109     {
110         vector<int> ans;
111         for(int i = 0; i < edges.size(); i++)
112         {
113             Edge& e = edges[i];
114             if(vis[e.from] && !vis[e.to] && e.cap > 0) ans.push_back(i);
115         }
116         return ans;
117     }
118 
119     void Reduce()
120     {
121         for(int i = 0; i < edges.size(); i++) edges[i].cap -= edges[i].flow;
122     }
123 };
124 
125 Dinic g;
126 
127 int main()
128 {
129     int n, e, c, kase = 0;
130     while(scanf("%d%d%d", &n, &e, &c) == 3 && n)
131     {
132         g.ClearAll(n);
133         while(e--)
134         {
135             int b1, b2, fp;
136             scanf("%d%d%d", &b1, &b2, &fp);
137             g.AddEdge(b1-1, b2-1, fp);
138         }
139         int flow = g.Maxflow(0, n-1);
140         printf("Case %d: ", ++kase);
141         if(flow >= c) printf("possible\n");
142         else
143         {
144             vector<int> cut = g.Mincut();
145             g.Reduce();
146             vector<Edge> ans;
147             for(int i = 0; i < cut.size(); i++)
148             {
149                 Edge& e = g.edges[cut[i]];
150                 e.cap = c;
151                 g.ClearFlow();
152                 if(flow + g.Maxflow(0, n-1) >= c) ans.push_back(e);
153                 e.cap = 0;
154             }
155             if(ans.empty()) printf("not possible\n");
156             else
157             {
158                 sort(ans.begin(), ans.end());
159                 printf("possible option:(%d,%d)", ans[0].from+1, ans[0].to+1);
160                 for(int i = 1; i < ans.size(); i++)
161                     printf(",(%d,%d)", ans[i].from+1, ans[i].to+1);
162                 printf("\n");
163             }
164         }
165     }
166     return 0;
167 }
View Code

 

数论小模板

 1 void div_fac(int m)
 2 {
 3     Len = 0;
 4     for(int i=2; i*i<=m; i++) if(m%i==0)
 5     {
 6         p[Len] = i;
 7         xc[Len] = 1;
 8         while(m%i==0) xc[Len] *= i , m /= i;
 9         Len++;
10     }
11     if(m>1) p[Len] = xc[Len++] = m;
12 }
13 
14 int getcount(int x,int p)
15 {
16     int ret = 0;
17     while(x)
18     {
19         x /= p;
20         ret += x;
21     }
22     return ret;
23 }
24 
25 int pow_mod(int a,int n,int mod)
26 {//快速幂
27     a%=mod;
28     int r = 1;
29     while(n)
30     {
31         if(n&1) r=(ll)r*a%mod;
32         a=(ll)a*a%mod;
33         n>>=1;
34     }
35     return r;
36 }
37 
38 int gcd(int a,int b,int &x,int &y)
39 {//最大公因数
40     if(b==0)
41     {
42         x=1;
43         y=0;
44         return a;
45     }
46     int d = gcd(b,a%b,y,x);
47     y-=a/b*x;
48     return d;
49 }
50 
51 int inv(int a,int mod)
52 {//求逆元
53     int x,y;
54     while( gcd(a,mod,x,y) != 1 );
55     return (x+mod)%mod;
56 }
57 
58 int crt(int a[],int m[],int n)
59 {//中国剩余定理
60     int M = 1;
61     for(int i=0; i<n; i++) M*=m[i];
62     int ret = 0;
63     for(int i=0; i<n; i++)
64     {
65         int x,y,tm=M/m[i];
66         gcd(tm,m[i],x,y);
67         ret=(ret+(ll)tm*x%M*a[i])%M;
68     }
69     return (ret+M)%M;
70 }
View Code

 

点hash

 1 const int maxn=500+10;
 2 const int prime = 100007;
 3 
 4 pii pt[maxn];
 5 
 6 struct point_hash
 7 {
 8     int _hash[prime+5];
 9     int _next[prime+5];
10     void clr()
11     {//清空hash表
12         cl(_next,0),cl(_hash,0);
13     }
14     void insert(pii pt,int index)
15     {//向hash表中插入标号是index的点pt
16         int x=pt.first,y=pt.second;
17         int h = (x*x+y*y) % prime;
18         int u = _hash[h];
19         while(u)
20         {
21             u=_next[u];
22         }
23         _next[index]=_hash[h];
24         _hash[h]=index;
25     }
26     bool ok(pii p)
27     {//查询hash表中点p是否存在
28         int x=p.first,y=p.second;
29         int h = (x*x+y*y) % prime;
30         int u = _hash[h];
31         while(u)
32         {
33             if(pt[u].first==x && pt[u].second==y)return true;
34             u=_next[u];
35         }
36         return false;
37     }
38 }pthash;
View Code

 

威尔逊定理

p为素数时 

 

从ACdream那里扒下来的

 

 1 #include <iostream>
 2 #include <string.h>
 3 #include <stdio.h>
 4 #include <math.h>
 5 
 6 using namespace std;
 7 typedef long long LL;
 8 
 9 int Work(LL n)
10 {
11     while(n % 4 == 0) n >>= 2;
12     if(n % 8 == 7) return 4;
13     LL i = 8, t = 9;
14     while(t <= n)
15     {
16         while(n % t == 0) n /= t;
17         i += 8;
18         t += i;
19     }
20     if(n == 1) return 1;
21     if(n % 2 == 0) n >>= 1;
22     if(n % 4 == 3) return 3;
23     LL k = 3;
24     while(k * k <= n)
25     {
26         if(n % k == 0) return 3;
27         k += 4;
28     }
29     return 2;
30 }
31 
32 int main()
33 {
34     LL n;
35     while(cin>>n)
36         cout<<Work(n)<<endl;
37     return 0;
38 }
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posted @ 2017-09-10 22:54  良将ℓ  阅读(444)  评论(0)  编辑  收藏  举报