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矩阵快速幂。先要处理出第i列每个状态下,让该状态填满,下一列可以出现的状态。因为N较大,可以矩阵加速。
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector> #include<map> #include<set> #include<queue> #include<stack> #include<iostream> using namespace std; typedef long long LL; const double pi=acos(-1.0),eps=1e-8; void File() { freopen("D:\\in.txt","r",stdin); freopen("D:\\out.txt","w",stdout); } inline int read() { char c = getchar(); while(!isdigit(c)) c = getchar(); int x = 0; while(isdigit(c)) { x = x * 10 + c - ‘0‘; c = getchar(); } return x; } LL n,MOD; struct Matrix { long long A[18][18]; int R, C; Matrix operator*(Matrix b); }; Matrix X, Y, Z; Matrix Matrix::operator*(Matrix b) { Matrix c; memset(c.A, 0, sizeof(c.A)); int i, j, k; for (i = 1; i <= R; i++) for (j = 1; j <= b.C; j++) for (k = 1; k <= C; k++) c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j]) % MOD) % MOD; c.R = R; c.C = b.C; return c; } void dfs(int a,int b,int t,int c) { if(t==4) { X.A[c+1][b+1]=1; return; } if(a&(1<<t)) dfs(a,b,t+1,c); else { dfs(a|(1<<t),b|(1<<t),t+1,c); if(t+1<4&&(a&(1<<(t+1)))==0) dfs(a+(1<<t)+(1<<(t+1)),b,t+1,c); } } void init() { memset(X.A, 0, sizeof X.A); memset(Y.A, 0, sizeof Y.A); memset(Z.A, 0, sizeof Z.A); Y.R = 16; Y.C = 16; for (int i = 1; i <= 16; i++) Y.A[i][i] = 1; X.R = 16; X.C = 16; for(int i=0;i<=15;i++) dfs(i,0,0,i); Z.R = 1; Z.C = 16; Z.A[1][1]=1; } void work() { while (n) { if (n % 2 == 1) Y = Y*X; n = n >> 1; X = X*X; } Z = Z*Y; printf("%lld\n",Z.A[1][16]); } int main() { while(~scanf("%lld%lld",&n,&MOD)) { if(n==0&&MOD==0) break; n++; init(); work(); } return 0; }
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原文地址:http://www.cnblogs.com/zufezzt/p/5740777.html