[Project Euler 520]Simbers 题解
传送门：https://projecteuler.net/problem=520
一个假数位dp。
矩阵乘法调了一年。。。
刚开始以为是个倍增fwt，发现怎么算复杂度都不对。
后面发现好像所有奇数和偶数本质是一样的，除了前导0要容斥一下。
似乎可以推式子了。
枚举$t$个奇数要用，$f_{i,j,k}$表示前i位，有$j$个偶数出现次数为偶数，有$k$个奇数出现次数为奇数
考虑$f_{i,j,k}$的贡献：
$f_{i,j,k}*(5-j) \to f_{i+1,j+1,k} $
$f_{i,j,k}*(t-k) \to f_{i+1,j,k+1}$
$f_{i,j,k}*j \to f_{i+1,j-1,k}$
$f_{i,j,k}*k \to f_{i+1,j,k-1}$
我们发现这是个线性递推，矩阵维护即可
调了一年。
//waz #include <bits/stdc++.h> using namespace std; #define mp make_pair #define pb push_back #define fi first #define se second #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((int)((x).size())) typedef pair<int, int> PII; typedef vector<int> VI; typedef long long int64; typedef unsigned int uint; typedef unsigned long long uint64; #define gi(x) ((x) = F()) #define gii(x, y) (gi(x), gi(y)) #define giii(x, y, z) (gii(x, y), gi(z)) int F() { char ch; int x, a; while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-'); if (ch == '-') ch = getchar(), a = -1; else a = 1; x = ch - '0'; while (ch = getchar(), ch >= '0' && ch <= '9') x = (x << 1) + (x << 3) + ch - '0'; return a * x; } const int mod = 1000000123; int inc(int a, int b) { a += b; return a >= mod ? a - mod : a; } int dec(int a, int b) { a -= b; return a < 0 ? a + mod : a; } struct mat { int a[50][50]; int n; void clear() { n = 0; memset(a, 0, sizeof a); } friend mat operator * (const mat &a, const mat &b) { mat c; c.clear(); c.n = a.n; for (int i = 1; i <= c.n; ++i) for (int j = 1; j <= c.n; ++j) for (int k = 1; k <= c.n; ++k) c.a[i][j] = inc(c.a[i][j], 1LL * a.a[i][k] * b.a[k][j] % mod); return c; } friend mat operator + (mat a, mat b) { for (int i = 1; i <= a.n; ++i) for (int j = 1; j <= b.n; ++j) a.a[i][j] = inc(a.a[i][j], b.a[i][j]); return a; } friend mat operator - (mat a, mat b) { for (int i = 1; i <= a.n; ++i) for (int j = 1; j <= b.n; ++j) a.a[i][j] = dec(a.a[i][j], b.a[i][j]); return a; } } g; int work(int x, int p, int t, int gg) { static int id[10][10]; int idt = 0; for (int i = 0; i <= p; ++i) for (int j = 0; j <= t; ++j) id[i][j] = ++idt; g.clear(); g.n = idt; for (int j = 0; j <= p; ++j) for (int k = 0; k <= t; ++k) { if (j + 1 <= p) { int x = id[j][k], y = id[j + 1][k]; g.a[x][y] = p - j; } if (k + 1 <= t) { int x = id[j][k], y = id[j][k + 1]; g.a[x][y] = t - k; } if (j - 1 >= 0) { int x = id[j][k], y = id[j - 1][k]; g.a[x][y] = j; } if (k - 1 >= 0) { int x = id[j][k], y = id[j][k - 1]; g.a[x][y] = k; } } /*for (int i = 1; i <= idt; ++i) for (int j = 1; j <= idt; ++j) printf("%d%c", g.a[i][j], " \n"[j == idt]); cerr << id[p][0] << ", " << id[p][t] << endl;*/ mat ret; ret.clear(); ret.n = idt; ret.a[1][id[p][gg]] = 1; for (; x; x >>= 1) { if (x & 1) ret = ret * g; g = g * g; } return ret.a[1][id[p][t]]; } int solve1(int x, int p, int t, int gg) { static int id[10][10]; int idt = 0; for (int i = 0; i <= p; ++i) for (int j = 0; j <= t; ++j) id[i][j] = ++idt; g.clear(); g.n = idt; for (int j = 0; j <= p; ++j) for (int k = 0; k <= t; ++k) { if (j + 1 <= p) { int x = id[j][k], y = id[j + 1][k]; g.a[x][y] = p - j; } if (k + 1 <= t) { int x = id[j][k], y = id[j][k + 1]; g.a[x][y] = t - k; } if (j - 1 >= 0) { int x = id[j][k], y = id[j - 1][k]; g.a[x][y] = j; } if (k - 1 >= 0) { int x = id[j][k], y = id[j][k - 1]; g.a[x][y] = k; } } mat ret, c; ret.clear(); c.clear(); c.n = idt; ret.n = idt; ret.a[1][id[p][gg]] = 1; static mat f[50], tt[50]; f[0] = g; tt[0] = g; for (int i = 1; i <= x; ++i) { f[i].clear(); f[i] = f[i - 1] * f[i - 1]; tt[i] = f[i - 1] * tt[i - 1] + tt[i - 1]; } for (int i = 0; i <= x; ++i) c = c + tt[i]; ret = ret * c; return ret.a[1][id[p][t]]; } int solve2(int x, int p, int t, int gg) { static int id[10][10]; int idt = 0; for (int i = 0; i <= p; ++i) for (int j = 0; j <= t; ++j) id[i][j] = ++idt; g.clear(); g.n = idt; for (int j = 0; j <= p; ++j) for (int k = 0; k <= t; ++k) { if (j + 1 <= p) { int x = id[j][k], y = id[j + 1][k]; g.a[x][y] = p - j; } if (k + 1 <= t) { int x = id[j][k], y = id[j][k + 1]; g.a[x][y] = t - k; } if (j - 1 >= 0) { int x = id[j][k], y = id[j - 1][k]; g.a[x][y] = j; } if (k - 1 >= 0) { int x = id[j][k], y = id[j][k - 1]; g.a[x][y] = k; } } mat ret, c; ret.clear(); c.clear(); c.n = idt; ret.n = idt; ret.a[1][id[p][gg]] = 1; static mat f[50], tt[50], pp[50]; f[0] = g; tt[0] = g; pp[0].clear(); pp[0].n = idt; for (int i = 1; i <= x; ++i) { f[i].clear(); f[i] = f[i - 1] * f[i - 1]; tt[i] = f[i - 1] * tt[i - 1] + tt[i - 1]; pp[i] = tt[i] - f[i]; } for (int i = 0; i <= x; ++i) c = c + pp[i]; ret = ret * c; return ret.a[1][id[p][t]]; } int C[110][110]; int work(int m) { int ans = 0; for (int n = 1; n <= m; ++n) for (int t = 0; t <= 5; ++t) { int x = dec(work(n, 5, t, 0), work(n - 1, 4, t + 1, 0)); ans = (ans + 1LL * x * C[5][t]) % mod; } return ans; } int main() { for (int i = *C[0] = 1; i <= 100; ++i) for (int j = C[i][0] = 1; j <= i; ++j) C[i][j] = inc(C[i - 1][j - 1], C[i - 1][j]); int m; int c = 39; /*int ret = 0; for (int i = 0; i <= 3; ++i) { ret = inc(ret, work(1 << i)); } cerr << ret << endl;*/ int ans = 0; for (int t = 0; t <= 5; ++t) ans = (ans + 1LL * solve1(c, 5, t, 0) * C[5][t]) % mod; for (int t = 0; t <= 5; ++t) ans = (ans - 1LL * solve2(c, 4, t + 1, 0) * C[5][t]) % mod; ans = dec(ans, work(1)); cout << ans << endl; return 0; }
