传送门:【HDU】5152 A Strange Problem
题目分析:本题的难点全在type=2上了,首先我们需要知道一个结论(为什么说是结论。。因为我不会推导T T而且在做这题之前我压根就不知道有这个结论!):A^x = A ^ (x%Phi(C) + Phi(C)) (mod C) (x >= Phi(C))。其中phi(C)是C的欧拉函数。
然后我们发现phi(C)一层套一层,最多到第18层就变成1了,所以我们只要对每个数保存其每一层的数,然后每次type=2时暴力修改该节点的值,大于18层直接截取到18层就好了,再往下都是1了。由于一共最多50000个type=2,所以邻接表最多开其一倍的大小(一开始的数也要插入)即可。
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std ; typedef long long LL ; #define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i ) #define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i ) #define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i ) #define clr( a , x ) memset ( a , x , sizeof a ) #define cpy( a , x ) memcpy ( a , x , sizeof a ) #define ls ( o << 1 ) #define rs ( o << 1 | 1 ) #define lson ls , l , m #define rson rs , m + 1 , r #define rt o , l , r #define root 1 , 1 , n #define mid ( ( l + r ) >> 1 ) const int MAXN = 100005 ; const int MAXE = 100005 ; const int mod = 2333333 ; struct Edge { LL v ; int n ; Edge () {} Edge ( LL v , int n ) : v ( v ) , n ( n ) {} } ; int mo[19] = { 2333333 , 2196720 , 580608 , 165888 , 55296 , 18432 , 6144 , 2048 , 1024 , 512 , 256 , 128 , 64 , 32 , 16 , 8 , 4 , 2 , 1 } ; Edge E[MAXE] ; int H[MAXN] , cntE ; int sum[MAXN << 2] ; int len[MAXN << 2] ; LL add[MAXN << 2] ; int n , m ; void clear () { cntE = 0 ; clr ( H , -1 ) ; } void addedge ( int u , LL v ) { E[cntE] = Edge ( v , H[u] ) ; H[u] = cntE ++ ; } int pow ( int a , int b , int mod ) { int res = 1 ; while ( b ) { if ( b & 1 ) res = ( LL ) res * a % mod ; a = ( LL ) a * a % mod ; b >>= 1 ; } return res % mod ; } void pushdown ( int o ) { if ( add[o] ) { add[ls] += add[o] ; add[rs] += add[o] ; sum[ls] = ( sum[ls] + add[o] % mod * len[ls] ) % mod ; sum[rs] = ( sum[rs] + add[o] % mod * len[rs] ) % mod ; add[o] = 0 ; } } void build ( int o , int l , int r ) { add[o] = 0 ; len[o] = r - l + 1 ; if ( l == r ) { scanf ( "%d" , &sum[o] ) ; addedge ( l , sum[o] ) ; sum[o] %= mod ; return ; } int m = mid ; build ( lson ) ; build ( rson ) ; sum[o] = ( sum[ls] + sum[rs] ) % mod ; } void update ( int L , int R , int v , int o , int l , int r ) { if ( L <= l && r <= R ) { add[o] += v ; sum[o] = ( sum[o] + ( LL ) len[o] * v ) % mod ; return ; } int m = mid ; pushdown ( o ) ; if ( L <= m ) update ( L , R , v , lson ) ; if ( m < R ) update ( L , R , v , rson ) ; sum[o] = ( sum[ls] + sum[rs] ) % mod ; } int calc ( int i , int x ) { if ( x == 19 ) return 0 ; LL tmp = E[i].n == -1 ? E[i].v : pow ( 2 , calc ( E[i].n , x + 1 ) , mo[x] ) + E[i].v ; return tmp < mo[x] ? tmp : tmp % mo[x] + mo[x] ; } void modify ( int x , int o , int l , int r ) { if ( l == r ) { E[H[l]].v += add[o] ; add[o] = 0 ; addedge ( l , 0 ) ; sum[o] = calc ( H[l] , 0 ) % mod ; return ; } int m = mid ; pushdown ( o ) ; if ( x <= m ) modify ( x , lson ) ; else modify ( x , rson ) ; sum[o] = ( sum[ls] + sum[rs] ) % mod ; } int query ( int L , int R , int o , int l , int r ) { if ( L <= l && r <= R ) return sum[o] ; int m = mid ; pushdown ( o ) ; if ( R <= m ) return query ( L , R , lson ) ; if ( m < L ) return query ( L , R , rson ) ; return ( query ( L , R , lson ) + query ( L , R , rson ) ) % mod ; } void solve () { int op , l , r , x ; clear () ; build ( root ) ; while ( m -- ) { scanf ( "%d" , &op ) ; if ( op == 1 ) { scanf ( "%d%d" , &l , &r ) ; printf ( "%d\n" , query ( l , r , root ) ) ; } else if ( op == 2 ) { scanf ( "%d" , &x ) ; modify ( x , root ) ; } else { scanf ( "%d%d%d" , &l , &r , &x ) ; update ( l , r , x , root ) ; } } } int main () { while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ; return 0 ; }