矩陣加速

• 數學作業
• • description
  • solution
  • code
• Arc of Dream
• • description
  • solution
  • code

數學作業

description

solution

$dp$狀態轉移方程，$d{p}_i=d{p}_{i?1}?10^{le{n}_i}+i(\mathrm{m}\mathrm{o}\mathrm{d}M)$

$n$巨大，分段矩陣加速
$$\left[\begin{array}{ccc}{f}_{i?1}& i?1& 1\end{array}\right]\times ?\begin{array}{ccc}10^{len}& 0& 0\\ 1& 1& 0\\ 1& 1& 1\end{array}?=\left[\begin{array}{ccc}{f}_i& i& 1\end{array}\right]$$

code

#include <cmath> #include <cstdio> #include <cstring> #define int long long int n, mod; int mi[20]; struct matrix {int n, m;int c[3][3];matrix() {n = m = 0;memset( c, 0, sizeof( c ) );}matrix operator * ( matrix &t ) {matrix ans;ans.n = n, ans.m = t.m;for( int i = 0;i <= n;i ++ )for( int j = 0;j <= t.m;j ++ )for( int k = 0;k <= m;k ++ )ans.c[i][j] = ( ans.c[i][j] + c[i][k] * t.c[k][j] % mod ) % mod;return ans;} }fast, ret;matrix qkpow( matrix x, int y ) {if( ! y ) return x;matrix ans;ans.n = ans.m = 2;for( int i = 0;i <= 2;i ++ )ans.c[i][i] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans; }signed main() {scanf( "%lld %lld", &n, &mod );mi[0] = 1;for( int i = 1;i <= 18;i ++ )mi[i] = mi[i - 1] * 10;ret.n = 0, ret.m = 2, ret.c[0][2] = 1;fast.n = fast.m = 2;for( int i = 1;;i ++ ) {memset( fast.c, 0, sizeof( fast.c ) );fast.c[0][0] = mi[i] % mod;fast.c[1][0] = fast.c[1][1] = fast.c[2][0] = fast.c[2][1] = fast.c[2][2] = 1;if( pow( 10, i ) <= n )fast = qkpow( fast, mi[i - 1] * 9 );elsefast = qkpow( fast, n - mi[i - 1] + 1 );ret = ret * fast;if( mi[i] > n ) break;}printf( "%lld\n", ret.c[0][0] % mod );return 0; }

Arc of Dream

description

solution

$AoD(n)=AoD(n?1)+a_{n?1}{b}_{n?1}=AoD(n?1)+(a_{n?2}?AX+AY)(b_{n?2}?BX+BY)$

$=AoD(n?1)+a_{n?2}{b}_{n?2}AXBX+a_{n?2}AXBY+b_{n?2}BXAY+AYBY$

滿滿的矩陣味道，赤裸裸的系數暗示

$$\begin{gathered} \left[\begin{array}{ccccccc}AoD(i)& a_{i?1}{b}_{i?1}& a_{i?1}& b_{i?1}& AYBY& AY& BY\end{array}\right] \\ \times \\ ?\begin{array}{ccccccc}1& 0& 0& 0& 0& 0& 0\\ AXBX& AXBX& 0& 0& 0& 0& 0\\ AX& AX& AX& 0& 0& 0& 0\\ BX& BX& 0& BX& 0& 0& 0\\ 1& 1& 0& 0& 1& 0& 0\\ 1& 1& 1& 0& 0& 1& 0\\ 1& 1& 0& 1& 0& 0& 1\end{array}? \\ || \\ \left[\begin{array}{ccccccc}AoD(i+1)& a_i{b}_i& a_i& b_i& AYBY& AY& BY\end{array}\right] \end{gathered}$$

code

#include <cstdio> #include <cstring> #define int long long #define mod 1000000007 struct matrix {int n, m;int c[8][8];matrix() {n = m = 0;memset( c, 0, sizeof( c ) );}matrix operator * ( matrix &t ) {matrix ans;ans.n = n, ans.m = t.m;for( int i = 1;i <= n;i ++ )for( int j = 1;j <= t.m;j ++ )for( int k = 1;k <= m;k ++ )ans.c[i][j] = ( ans.c[i][j] + c[i][k] * t.c[k][j] ) % mod;return ans;} }g, ret;matrix qkpow( matrix x, int y ) {matrix ans;ans.n = x.n, ans.m = x.m;for( int i = 1;i <= ans.n;i ++ )ans.c[i][i] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans; }signed main() {int n, A0, AX, AY, B0, BX, BY;while( ~ scanf( "%lld %lld %lld %lld %lld %lld %lld", &n, &A0, &AX, &AY, &B0, &BX, &BY ) ) {g.n = g.m = ret.m = 7, ret.n = 1;if( ! n ) {printf( "0\n" );continue;}if( n == 1 ) {printf( "%lld\n", A0 * B0 % mod );continue;}memset( g.c, 0, sizeof( g.c ) );memset( ret.c, 0, sizeof( ret.c ) );ret.c[1][1] = ret.c[1][2] = A0 * B0 % mod;ret.c[1][3] = A0, ret.c[1][4] = B0;ret.c[1][5] = AY * BY % mod;ret.c[1][6] = AY, ret.c[1][7] = BY;g.c[1][1] = g.c[5][1] = g.c[5][2] = g.c[6][3] = g.c[7][4] = g.c[5][5] = g.c[6][6] = g.c[7][7] = 1;g.c[2][1] = g.c[2][2] = AX * BX % mod;g.c[3][1] = g.c[3][2] = AX * BY % mod;g.c[4][1] = g.c[4][2] = AY * BX % mod;g.c[3][3] = AX, g.c[4][4] = BX;g = qkpow( g, n - 1 );ret = ret * g;printf( "%lld\n", ret.c[1][1] );}return 0; }

HH去散步

之前已經寫過博客了，不再贅述

總結

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