51Nod-1197-字符串的数量 V2
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ACM模版
描述
题解
这道题好阔怕，在V1的讨论区有最牛逼的题解，Orz！！！
虽然有题解，有代码，可是由于代数还没有学，所以对部分有些懵逼，先Mark一下，过一阵子能力有所提升后好好瞅瞅喽……
提供两份代码，对于这道题，代码One和代码Two效率差距不明显，代码Two相比代码One多了一个预处理，针对V3的多组数据可以提高很多效率！！！
代码
One：
#include <stdio.h>typedef long long ll;const int MAXN = 61;const int MOD = 1e9 + 7;int deg;int iact[MAXN + 1];int val[MAXN + 1];int nxt[MAXN + 1];int seq[MAXN + 1];ll n, m;inline int mod_add(int x, int y){ return x + y < MOD ? x + y : x + y - MOD;}inline int mod_dec(int x, int y){ return x - y >= 0 ? x - y : x - y + MOD;}int value(int a){ if (a < deg) { return val[a]; } static int pre[MAXN], suf[MAXN]; pre[0] = a; for (int i = 1; i < deg; ++i) { pre[i] = (ll)(a - i) * pre[i - 1] % MOD; } suf[deg - 1] = a - deg + 1; for (int i = deg - 2; i >= 0; --i) { suf[i] = (ll)(a - i) * suf[i + 1] % MOD; } int ret = 0; for (int i = 0; i < deg; ++i) { int tmp = (ll)val[i] * iact[i] % MOD * iact[deg - 1 - i] % MOD; if (i > 0) { tmp = (ll)tmp * pre[i - 1] % MOD; } if (i < deg - 1) { tmp = (ll)tmp * suf[i + 1] % MOD; } if (((deg - 1 - i) & 1) && tmp > 0) { tmp = MOD - tmp; } ret += tmp; if (ret >= MOD) { ret -= MOD; } } return ret;}struct Matrix{ int r, c, num[MAXN][MAXN]; Matrix operator * (const Matrix &t) const { Matrix ret = {r, t.c}; for (int i = 0; i < r; ++i) { for (int j = 0; j < c; ++j) { for (int k = 0; k < t.c; ++k) { ret.num[i][k] = (ret.num[i][k] + (ll)num[i][j] * t.num[j][k]) % MOD; } } } return ret; } Matrix pow(ll k) { Matrix ret = {r, r}, tmp = *this; for (int i = 0; i < r; ++i) { ret.num[i][i] = 1; } for (; k > 0; k >>= 1, tmp = tmp * tmp) { if (k & 1) { ret = ret * tmp; } } return ret; }} A;int main(){// int T;// scanf("%d", &T);// while (T--)// { iact[0] = 0; iact[1] = 1; for (int i = 2; i <= MAXN; ++i) { iact[i] = MOD - MOD / i * (ll)iact[MOD % i] % MOD; } iact[0] = 1; for (int i = 1; i <= MAXN; ++i) { iact[i] = iact[i - 1] * (ll)iact[i] % MOD; } int len; scanf("%lld%lld", &n, &m); ll tmp = (n << 1) - 1; deg = val[0] = 1; for (len = 1; --tmp >= 0; ++len) { for (int i = 0, pos = tmp & 1; i <= deg; ++i, pos = mod_add(pos, 2)) { nxt[i] = mod_add(i > 0 ? nxt[i - 1] : 0, value(pos)); } ++deg; for (int i = 0; i < deg; ++i) { val[i] = nxt[i]; } tmp >>= 1; seq[len] = value(tmp % MOD); } --len; ll tmp_ = n - 1; deg = val[0] = 1; for (int j = 1; --tmp_ >= 0; ++j) { for (int i = 0, pos = tmp_ & 1; i <= deg; ++i, pos = mod_add(pos, 2)) { nxt[i] = mod_add(i > 0 ? nxt[i - 1] : 0, value(pos)); } ++deg; for (int i = 0; i < deg; ++i) { val[i] = nxt[i]; } tmp_ >>= 1; seq[j] = mod_dec(seq[j], value(tmp_ % MOD)); } A = (Matrix){len, len}; for (int i = 1; i < len; ++i) { A.num[i][i - 1] = 1; } for (int i = 0; i < len; ++i) { A.num[i][len - 1] = seq[len - i]; } A = A.pow(m); printf("%d\n", A.num[len - 1][len - 1]);// } return 0;}Two：
#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <cmath>using namespace std;typedef long long ll;const int MOD = 1e9 + 7;const int MAXN = 63;void add(ll &x, ll y){ x += y; if (x >= MOD) { x -= MOD; }}ll i, j;ll p, k;ll MAXP;ll A[MAXN + 5][MAXN + 5];ll E[MAXN + 5];ll pr[20005];ll MA[20005];vector<ll> v[MAXN];vector<ll> va[MAXN];vector<ll> vb[MAXN];vector<ll> vr[MAXN];ll modAdd(ll a, ll b){ a += b; return a >= MOD ? (a - MOD) : a;}ll modMul(ll a , ll b){ a *= b; return a >= MOD ? (a % MOD) : a;}void multiply(const vector<ll> &Q, vector<ll> &Qr){ int size = (int)Qr.size(); for (int i = 0 ; i < size; i++) { Qr[i] = 0; for (int j = max(0 , 2 * i - size + 1); j < min(size, (2 * i + 1)); j++) { ll x = (j & 1) ? ((MOD - Q[j])) : (Q[j]); ll y = modMul(x, Q[2 * i - j]); Qr[i] = modAdd(Qr[i], y); } }}void cal(ll *A, ll n, ll k, ll *C){ vector<ll> Q(k + 3); vector<ll> Qr(k + 3); n--; while (n >= k) { Q[0] = 1; for (ll i = 0; i < k; i++) { Q[i + 1] = -C[i]; if (Q[i + 1] < 0) { Q[i + 1] += MOD; } } for (ll i = k; i < (2 * k); i++) { A[i] = 0; for (ll j = 0; j < k; j++) { ll add = modMul(A[i - 1 - j], C[j]); A[i] = modAdd(A[i], add); } } multiply(Q, Qr); for (ll i = 0; i < k; i++) { C[i] = (-Qr[i + 1]); if (C[i] < 0) { C[i] += MOD; } } int offset = n & 1; for (ll i = 0; i < k; i++) { A[i] = A[2 * i + offset]; } n /= 2; } printf("%lld\n", A[n]);}int power(ll x, ll y){ int sum = 1; for (; y; y >>= 1) { if (y & 1) { sum = 1ll * sum * x % MOD; } x = 1ll * x * x % MOD; } return sum;}void guass(ll n){ for (int i = 1; i <= n; ++i) { int j; for (j = i; j <= n; ++j) { if (A[j][i]) { break; } } for (int k = 1; k <= n + 1; ++k) { swap(A[i][k], A[j][k]); } int p = power(A[i][i], MOD - 2); for (int j = i + 1; j <= n; ++j) { if (A[j][i]) { int now = 1ll * A[j][i] * p % MOD; for (int k = i; k <= n + 1; ++k) { add(A[j][k], MOD - 1ll * A[i][k] * now % MOD); } } } } for (ll i = n; i; --i) { for (ll j = i + 1; j <= n; ++j) { add(A[i][n + 1], MOD - 1ll * A[i][j] * E[j] % MOD); } E[i] = 1ll * A[i][n + 1] * power(A[i][i], MOD - 2) % MOD; }}long long n, m, sA[MAXN], sB[MAXN];long long Count(vector<ll> A, ll R){ ll i, j, sum = 0; R %= MOD; for (i = 0; i < A.size(); ++i) { ll S = 0, now = 1; for (j = 0; j < vr[i].size(); ++j) { add(S, 1ll * vr[i][j] * now % MOD), now = 1ll * now * R % MOD; } add(sum, 1ll * S * A[i] % MOD); } return sum;}int main(){ for (i = 0; i < MAXN; ++i) { memset(A, 0, sizeof(A)); for (j = 1; j <= i + 2; ++j) { for (k = 1; k <= i + 2; ++k) { A[j][k] = power(j, k - 1); } A[j][i + 3] = power(2 * j - 1, i); add(A[j][i + 3], A[j - 1][i + 3]); if (j > 1) { add(A[j][i + 3], power(2 * j - 2, i)); } } guass(i + 2); for (j = 1; j <= i + 2; ++j) { v[i].push_back(E[j]); } } for (i = 0; i < MAXN; ++i) { memset(A, 0, sizeof(A)); for (j = 1; j <= i + 2; ++j) { for (k = 1; k <= i + 2; ++k) { A[j][k] = power(j, k - 1); } A[j][i + 3] = power(j, i); (A[j][i + 3] += A[j - 1][i + 3]) %= MOD; } guass(i + 2); for (j = 1; j <= i + 2; ++j) { vr[i].push_back(E[j]); } }// int T;// scanf("%d", &T);// for (; T--; )// { scanf("%lld%lld", &n, &m); MAXP = (int)(log2(n) + 1); memset(sA, 0, sizeof(sA)); memset(sB, 0, sizeof(sB)); for (i = 1; i <= MAXP; ++i) { va[i].clear(), vb[i].clear(); } sA[1] = n / 2; sB[1] = n; vb[1].push_back(1); for (i = 2; i <= MAXP; ++i) { sA[i] = sA[i - 1] / 2; sB[i] = sB[i - 1] / 2; long long str = Count(vb[i - 1], sB[i - 1]); for (j = 0; j < i; ++j) { vb[i].push_back(0), va[i].push_back(0); } vb[i][0] = str; for (j = 0; j < vb[i - 1].size(); ++j) { ll wei = (MOD - vb[i - 1][j]); for (k = 0; k < (int)v[j].size(); ++k) { add(vb[i][k], 1ll * wei * v[j][k] % MOD); } } ll End = Count(vb[i - 1], sA[i - 1]); va[i][0] = (str - End + MOD) % MOD; add(va[i][0], Count(va[i - 1], sA[i - 1])); for (j = 0; j < va[i - 1].size(); ++j) { ll wei = (MOD - va[i - 1][j]); for (k = 0; k < v[j].size(); ++k) { add(va[i][k], 1ll * wei * v[j][k] % MOD); } } } for (i = 1; i <= MAXP; ++i) { pr[i] = Count(va[i], sA[i]); add(pr[i], Count(vb[i], sB[i])); add(pr[i], MOD - Count(vb[i], sA[i])); } memset(MA, 0, sizeof(MA)); MA[0] = 1; for (i = 1; i < MAXP; ++i) { for (j = 1; j <= i; ++j) { add(MA[i], 1ll * MA[i - j] * pr[j] % MOD); } } for (i = 0; i < MAXP; ++i) { swap(pr[i], pr[i + 1]); } if (m < MAXP) { printf("%lld\n", MA[m]); } else { cal(MA, m + 1, MAXP, pr); }// } return 0;}参考
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