数学作业

solution

$d p$ 状态转移方程， $dpi=dpi−1∗10leni+i(modM)dp_{i}=dp_{i-1}*10^{len_i}+i\pmod M$

$n$ 巨大，分段矩阵加速
$[fi−1i−11]×[10len00110111]=[fii1]\begin{bmatrix} f_{i-1}&i-1&1 \end{bmatrix} \times \begin{bmatrix} 10^{len}&0&0\\ 1&1&0\\ 1&1&1 \end{bmatrix}= \begin{bmatrix} f_{i}&i&1 \end{bmatrix}$

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

solution

$AoD(n)=AoD(n-1)+a_{n-1}b_{n-1}=AoD(n-1)+(a_{n-2}*AX+AY)(b_{n-2}*BX+BY)$

$AoD(n-1)+a_{n-2}b_{n-2}AXBX+a_{n-2}AXBY+b_{n-2}BXAY+AYBY$

满满的矩阵味道，赤裸裸的系数暗示
在这里插入图片描述

$[AoD(i)ai−1bi−1ai−1bi−1AYBYAYBY]×[1000000AXBXAXBX00000AXAXAX0000BXBX0BX000110010011100101101001]∣∣[AoD(i+1)aibiaibiAYBYAYBY]\begin{bmatrix} AoD(i)&a_{i-1}b_{i-1}&a_{i-1}&b_{i-1}&AYBY&AY&BY \end{bmatrix} \\ \times \\ \begin{bmatrix} 1&0&0&0&0&0&0\\ AXBX&AXBX&0&0&0&0&0\\ AX&AX&AX&0&0&0&0\\ BX&BX&0&BX&0&0&0\\ 1&1&0&0&1&0&0\\ 1&1&1&0&0&1&0\\ 1&1&0&1&0&0&1\\ \end{bmatrix} \\ || \\ \begin{bmatrix} AoD(i+1)&a_ib_i&a_i&b_i&AYBY&AY&BY \end{bmatrix}$

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;
}