标签:line scan span clu hid one 技术 技术分享 close
http://www.lydsy.com/JudgeOnline/problem.php?id=2004
矩阵快速幂。。。并没有想出来。。。
理解地不是很好 先挖个坑
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int N = 210, mod = 30031; struct mat { int a[N][N]; } x, A; int n, k, p, cnt; int st[N]; mat operator * (mat A, mat B) { mat ret; memset(ret.a, 0, sizeof(ret.a)); for(int i = 1; i <= cnt; ++i) for(int j = 1; j <= cnt; ++j) for(int k = 1; k <= cnt; ++k) ret.a[i][j] = (ret.a[i][j] + A.a[i][k] * B.a[k][j]) % mod; return ret; } mat power(mat x, int t) { mat ret; memset(ret.a, 0, sizeof(ret.a)); for(int i = 1; i <= cnt; ++i) ret.a[i][i] = 1; for(; t; t >>= 1, x = x * x) if(t & 1) ret = ret * x; return ret; } void dfs(int d, int pos, int num) { if(d == k) { st[++cnt] = num; return; } for(int i = pos; i; --i) dfs(d + 1, i - 1, num | (1 << (i - 1))); } int main() { scanf("%d%d%d", &n, &k, &p); dfs(1, p - 1, 1 << (p - 1)); for(int i = 1; i <= cnt; ++i) for(int j = 1; j <= cnt; ++j) { int s = st[i] ^ (st[j] << 1) ^ (1 << p); if(s == (s & -s)) x.a[i][j] = 1; } x = power(x, n - k); A.a[1][1] = 1; A = x * A; printf("%d\n", A.a[1][1]); return 0; }
标签:line scan span clu hid one 技术 技术分享 close
原文地址:http://www.cnblogs.com/19992147orz/p/6828212.html