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polya定理(涂的颜色有数量限制)注：calcul函数一定情况下可通用。#include"bits/stdc++.h" using namespace std; typedef unsigned long long LL; int sum,a[3],b[3]; LL c[41][41]; int gcd(int a,int b){ if(b == 0) return a; re...

Let it Bead Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6075   Accepted: 4068 Description "Let it Bead" company is located upstairs at 700 Cannery Row in

大数模板+Polya定理 Count the Tetris Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2681    Accepted Submission(s): 752 Problem Description

先算出总的状态转移方程：dp[i]  = 3*dp[i-1]-dp[i-2]+2; 与bzoj1002题转移方程相同。 因为n对于m可能无逆元，所以mod = n*m。可能会爆longlong。所以用类比于快速幂的写法，写快速乘。然后用矩阵乘法+欧拉函数优化。 Toy Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 327

参考题解 #include"bits/stdc++.h" using namespace std; typedef long long LL; const int MX = 1e3+7; const int mod = 1e9+7; int n,m,K; LL fac[MX], inv[MX]; LL ans = 1; void init() { int n = 1e3; f...