4105: [Thu Summer Camp 2015]平方运算

首先嘛这道题目只要知道一个东西就很容易了：所有循环的最小公约数<=60,成一条链的长度最大为11，那么我们就可以用一个很裸的方法。对于在链上的数，我们修改直接暴力找出并修改。对于在环上的数，我们对每一步建立一颗线段树，那么修改就变成了交换60棵线段树的某个子树。然后我们就可以愉快的写啦~~~

在想的时候被同学坑了，他说最长的循环不超过60.。。。

在实现方面，我用一个map+树状数组维护链上的答案，然后用60棵线段树来维护这个环。懂了之后还是很好写哒~~~

Code：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 const int N =  + , kCycle =  + ;

 int n, m, p, x[N];

 int dist[N], cycle = ;

 inline int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
 inline int LCM(int a, int b) { return a * b / GCD(a, b); }

 int pre[N];

 void Analysis(int a) {
   dist[a] = -;
   int b = a * a % p;
   if (dist[b] == -) {
     pre[b] = a;
     Analysis(b);
   } else if (dist[b] == -) {
     int cnt = ;
     for (int i = a; i != b; i = pre[i], ++cnt) dist[i] = ;
     dist[b] = ;
     cycle = LCM(cycle, cnt);
   }
   if (dist[a] == -) dist[a] = dist[b] + ;
 }

 inline int pos(int l, int r) { return (l + r) | (l != r); }

 int tree[N * ][kCycle], shift[N * ]; 

 inline void Merge(int u[], int a[], int b[]) {
   for (int i = ; i < cycle; ++i) u[i] = a[i] + b[i];
 }

 inline void Shift(int id, int v) {
   static int temp[kCycle * ];
   (shift[id] += v) %= cycle;
   for (int i = ; i < cycle; ++i) temp[i] = tree[id][(i + v) % cycle];
   std::copy(temp, temp + cycle, tree[id]);
 }

 void Modify(int l, int r, int u, int v) {
   int id = pos(l, r);
   for (int i = ; i < cycle; ++i, (v *= v) %= p) tree[id][i] += v;
   if (l == r) return;
   int mid = (l + r) / ;
   if (shift[id]) {
     Shift(pos(l, mid), shift[id]);
     Shift(pos(mid + , r), shift[id]);
     shift[id] = ;
   }
   if (u <= mid) Modify(l, mid, u, v); else Modify(mid + , r, u, v);
 }

 void Cycle(int l, int r, int u, int v) {
   int id = pos(l, r);
   if (u <= l && r <= v) {
     Shift(id, );
     return;
   }
   int mid = (l + r) / ;
   if (shift[id]) {
     Shift(pos(l, mid), shift[id]);
     Shift(pos(mid + , r), shift[id]);
     shift[id] = ;
   }
   if (u <= mid) Cycle(l, mid, u, v);
   if (v > mid) Cycle(mid + , r, u, v);
   Merge(tree[id], tree[pos(l, mid)], tree[pos(mid + , r)]);
 }

 int Query(int l, int r, int u, int v) {
   int id = pos(l, r);
   if (u <= l && r <= v) return tree[id][];
   int mid = (l + r) / , res = ;
   if (shift[id]) {
     Shift(pos(l, mid), shift[id]);
     Shift(pos(mid + , r), shift[id]);
     shift[id] = ;
   }
   if (u <= mid) res += Query(l, mid, u, v);
   if (v > mid) res += Query(mid + , r, u, v);
   return res;
 }

 class BIT {
   int data[N];

  public:
   inline void Add(int p, int v) { for (; p <= n; p += p & -p) data[p] += v; }
   inline int Query(int p) {
     int res = ;
     for (; p; p ^= p & -p) res += data[p];
     return res;
   }
   inline int Query(int l, int r) { return Query(r) - Query(l - ); }

 } cnt, sum;

 void Sqr(int l, int r) {
   if (!cnt.Query(l, r)) return;
   if (l == r) {
     cnt.Add(l, -);
     sum.Add(l, -x[l]);
     if (cnt.Query(l, l) == )
       Modify(, n, l, (x[l] *= x[l]) %= p);
     else
       sum.Add(l, (x[l] *= x[l]) %= p);
   } else {
     int mid = (l + r) / ;
     Sqr(l, mid);
     Sqr(mid + , r);
   }
 }

 int main() {
   scanf("%d%d%d", &n, &m, &p);
   for (int i = ; i <= n; ++i) scanf("%d", x + i);
   memset(dist, -, sizeof dist);
   for (int i = ; i < p; ++i) if (dist[i] == -) Analysis(i);
   for (int i = ; i <= n; ++i) {
     if (dist[x[i]]) {
       sum.Add(i, x[i]);
       cnt.Add(i, dist[x[i]]);
     } else {
       Modify(, n, i, x[i]);
     }
   }
   while (m--) {
     int op, l, r;
     scanf("%d%d%d", &op, &l, &r);
     switch (op) {
       case :
         Cycle(, n, l, r);
         Sqr(l, r);
         break;
       case :
         printf("%d\n", Query(, n, l, r) + sum.Query(l, r));
         break;
     }
   }
   return ;
 }