常用模板

• 1、辗转相除法求最大公约数

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int gcd(int a,int b) {     return b == 0?a:gcd(b,a%b); }

• 2、最大公约数

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int lcm(int a,int b) {     return a/gcd(a,b)*b; //先除后乘防止a*b溢出 }

• 3、拓展欧几里德算法

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void extgcd(__int64 a,__int64 b,__int64 &x,__int64 &y) {     if (!b)     {         x = 1;         y = 0;     }     else     {         extgcd(b,a%b,y,x);         y -= x*(a/b);     } }

• 4、求区间[m,n]内的素数（ 埃氏筛选法）

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int m = sqrt(n+0.5); memset(vis,0,sizeof(vis)); for (int i = 2; i <= m; i++) {     if (!vis[i])     {         for (int j = i*i; j <= n; j += i)         {             vis[j] = 1;         }     } }   版本二O(nloglogn) int p = 0; memset(is_prime,true,sizeof(is_prime)); is_prime[0] = is_prime[1] = false; for(int i = 2; i <= MAX; i++) {     if (is_prime[i])     {         prime[p++] = i;         for (int j = 2 * i; j <= MAX; j += i)         {             is_prime[j] = false;         }     } }

• 5、无平方因子的数

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1	对于不超过√m 的所有素数p，筛选区间[n,m]内p² 的所有倍数

• 6、快速乘

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LL mod_multi(LL x,LL y,LL mod)   //拆分成y个x相加 {     LL res = 0;     while (y)     {         if (y&1)         {             res += x;             while (res >= mod)  res -= mod;           }         x += x;         while (x >= mod)    x -= mod;         y >>= 1;     }     return res; }

• 7、快速幂

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LL mod_pow(LL x,LL n,LL mod) {     LL res = 1;     while (n > 0)     {         if (n&1)         {             res = mod_mulit(res,x,mod);             // res = res*x%mod;         }         x = mod_multi(x,x,mod);         //x = x*x%mod;         n >>= 1;     }     return res; }

• 8、二分查找

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// binsearch_min 返回key(可能有重复)第一次出现的下标，如无return -1   int binsearch_min(int * arr, int lef, int rig, int key) {     if(!arr)    return -1;     int left = lef, right = rig;     while(left < right)     {         int mid = left + ((right-left)>>1);         if(arr[mid] < key)         {             left = mid+1;         }         else         {             right = mid;         }     }     if(arr[left] == key)    return left;     return -1; }   // binsearch_max 返回key(可能有重复)最后一次出现的下标，如无return -1 int binsearch_max(int * arr, int lef, int rig, int key) {     if(!arr)    return -1;     int left = lef, right = rig;     while(left < right -1)     {         int mid = left + ((right-left)>>1);         if(arr[mid] <= key)         {             left = mid;         }         else         {             right = mid;         }     }     if(arr[right] == key) // 找max，先判断right     {         return right;     }     else if(arr[left] == key)     {         return left;     }     else         return -1; }

• 9、并查集

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void init() {     memset(father,0,sizeof(father));     memset(rk,0,sizeof(rk));     for (int i = 0; i <= N; i++)     {         father[i] = i;     } }   int find(int x) {     int r = x;     while (father[r] != r)     {         r = father[r];     }     int i = x,j;     while (i != r)           //路径压缩     {         j = father[i];         father[i] = r;         i = j;     }     return r; } /* int find(int x) {     return father[x] == x?father[x]:father[x] = find(father[x]); } */   void unite(int x,int y) {     int fx,fy;     fx = find(x);     fy = find(y);     if (fx == fy)   return;     if (rk[fx] < rk[fy])     {         father[fx] = fy;     }     else     {         father[fy] = fx;         if (rk[fx] == rk[fy])         {             rk[fx]++;         }     } }

• 10、生成1~n的排列

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void permutation(int n,int *A,int cur) {     if (cur == n)                   //递归边界     {         for (int i = 0;i < n;i++)    printf("%d ",A[i]);         printf("\n");     }     else for (int i = 1;i <= n;i++) //尝试在A[cur}中填各种整数i     {         int ok = 1;         for (int j = 0;j < cur;j++)  if (A[j] == i)  ok = 0;   //如果i已经在A[0]~A[cur-1]出现过，则不能再选         if (ok)         {             A[cur] = i;             permutation(n,A,cur+1); //递归调用         }     } }

• 11、生成可重集的排列

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void permutation(int n,int *P,int *A,int cur) //输入数组P，并按字典序输出数组A各元素的所有全排列 {     if (cur == n)     {         for (int i = 0; i < n; i++)  printf("%d ",A[i]);         printf("\n");     }     else for (int i = 0; i < n; i++)     {         if (!i || P[i] != P[i-1])         {             int c1 = 0,c2 = 0;             for (int j = 0; j < cur; j++)    if (A[j] == P[i])   c1++;             for (int j = 0; j < n; j++)  if (P[i] == P[j])   c2++;             if (c1 < c2)             {                 A[cur] = P[i];                 permutation(n,P,A,cur+1);             }         }     } }

• 12、强连通分量的targin算法

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/* 寻找有向图强连通分量的tarjan算法  * index表示的就是时间戳  * scc_cnt 表示强连通分量的个数  * belong[u] 表示结点u属于哪一个强连通分量  * inst[u] 表示结点u是否仍然在栈中  * st[] 和 top 分别表示栈和栈顶位置  *index = scc_cnt = top = 0*/ void targin(int u) {     int v;     dfn[u] = low[u] = ++index;     st[++top] = u;     inst[u] = 1;     for (int i = head[u];i != -1;i = edge[i].next)     {         v = edge[i].v;         if (!dfn[v])         {             targin(v);             low[u] = min(low[u],low[v]);         }         else if (inst[v])             low[u] = min(low[u],dfn[v]);     }     if (low[u] == dfn[u])     {         scc_cnt++;         do         {             v = st[top--];             inst[v] = 0;             belong[v] = scc_cnt;         }         while (u != v);     } }

• 13、SBT

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/* * tree[x].left   表示以 x 为节点的左儿子 * tree[x].right  表示以 x 为节点的右儿子 * tree[x].size   表示以 x 为根的节点的个数（大小） */       struct SBT {     int key,left,right,size; } tree[10010]; int root = 0,top = 0;    void left_rot(int &x)         // 左旋 {     int y = tree[x].right;     if (!y) return;     tree[x].right = tree[y].left;     tree[y].left = x;     tree[y].size = tree[x].size;     tree[x].size = tree[tree[x].left].size + tree[tree[x].right].size + 1;     x = y; }    void right_rot(int &x)        //右旋 {     int y = tree[x].left;     if (!y) return;     tree[x].left = tree[y].right;     tree[y].right = x;     tree[y].size = tree[x].size;     tree[x].size = tree[tree[x].left].size + tree[tree[x].right].size + 1;     x = y; }    void maintain(int &x,bool flag)  //维护SBT状态 {     if (!x) return;     if (flag == false)           //左边     {         if(tree[tree[tree[x].left].left].size > tree[tree[x].right].size)//左孩子的左孩子大于右孩子             right_rot(x);         else if (tree[tree[tree[x].left].right].size > tree[tree[x].right].size) //左孩子的右孩子大于右孩子         {             left_rot(tree[x].left);             right_rot(x);         }         else             return;     }     else                       //右边     {         if(tree[tree[tree[x].right].right].size > tree[tree[x].left].size)//右孩子的右孩子大于左孩子             left_rot(x);         else if(tree[tree[tree[x].right].left].size > tree[tree[x].left].size) //右孩子的左孩子大于左孩子         {             right_rot(tree[x].right);             left_rot(x);         }         else             return;     }     maintain(tree[x].left,false);     maintain(tree[x].right,true);     maintain(x,true);     maintain(x,false); }    void insert(int &x,int key)  //插入 {     if (x == 0)     {         x = ++top;         tree[x].left = 0;         tree[x].right = 0;         tree[x].size = 1;         tree[x].key = key;     }     else     {         tree[x].size++;         if(key < tree[x].key)             insert(tree[x].left,key);         else             insert(tree[x].right,key);//相同元素可插右子树         maintain(x,key >= tree[x].key);     } }    int remove(int &x,int key)  //利用后继删除 {     tree[x].size--;     if(key > tree[x].key)         remove(tree[x].right,key);     else  if(key < tree[x].key)         remove(tree[x].left,key);     else  if(tree[x].left !=0 && !tree[x].right) //有左子树，无右子树     {         int tmp = x;         x = tree[x].left;         return tmp;     }     else if(!tree[x].left && tree[x].right != 0) //有右子树，无左子树     {         int tmp = x;         x = tree[x].right;         return tmp;     }     else if(!tree[x].left && !tree[x].right)    //无左右子树     {         int tmp = x;         x = 0;         return tmp;     }     else                                      //左右子树都有     {         int tmp = tree[x].right;         while(tree[tmp].left)             tmp = tree[tmp].left;         tree[x].key = tree[temp].key;         remove(tree[x].right,tree[tmp].key);     } }    int getmin(int x)    //求最小值 {     while(tree[x].left)         x = tree[x].left;     return tree[x].key; }    int getmax(int x)    //求最大值 {     while(tree[x].right)         x = tree[x].right;     return tree[x].key; }    int pred(int &x,int y,int key)   //前驱，y初始前驱，从0开始, 最终要的是返回值的key值 {     if(x == 0)         return y;     if(key > tree[x].key)         return pred(tree[x].right,x,key);     else         return pred(tree[x].left,y,key); }    int succ(int &x,int y,int key)   //后继，同上 {     if(x == 0)         return y;     if(key < tree[x].key)         return succ(tree[x].left,x,key);     else         return succ(tree[x].right,y,key); }    int select(int &x,int k)   //查找第k小的数 {     int r = tree[tree[x].left].size + 1;     if(r == k)         return tree[x].key;     else if(r < k)         return select(tree[x].right,k - r);     else         return select(tree[x].left,k); }    int rank(int &x,int key)   //key排第几 {     if(key < tree[x].key)     {         return rank(tree[x].left,key);     }     else if(key > tree[x].key)         return rank(tree[x].right,key) + tree[tree[x].left].size + 1;     else         return tree[tree[x].left].size + 1; }

• 14、关闭同步

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1 2	ios::sync_with_stdio(false); cin.tie(NULL);