Magic Pairs

problem

已知$A_{0}x+B_{0}y\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$恒成立

求所有满足$Ax+By\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$成立的$(A,B)$数量，$0\le A,B<N$​

solution

既然$A_{0}x+B_{0}y\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$都已经成立了

那么$2A_{0}x+2B_{0}y\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$​也同样成立

换言之$k{A}_{0}x+k{B}_{0}y\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$​​都成立

无脑倍加，还对$A,B$划了范围，肯定是会有周期循环的

注意$A_0,B_0$要先模一下$n$​

code

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector < pair < int, int > > ans;int main() {int n, a0, b0;scanf( "%d %d %d", &n, &a0, &b0 );a0 %= n, b0 %= n;int a = a0, b = b0;do {ans.push_back( make_pair( a, b ) );a = ( a + a0 ) % n;b = ( b + b0 ) % n;}while( a != a0 || b != b0 );sort( ans.begin(), ans.end() );printf( "%d\n", ans.size() );for( int i = 0;i < ans.size();i ++ )printf( "%d %d\n", ans[i].first, ans[i].second );return 0;
}

CF107D Crime Management

problem

solution

设$li{m}_i:i$字符所有限制之积

所有cnt相乘小于等于123，换言之：如果枚举每个字符$i$的出现次数取模$li{m}_i$，哈希的总状态数不超过$123$

取模后的状态转移是固定的，现在$i$出现了$j$次，那么下一个状态$(j+1)\%li{m}_i$就可以设为$1$​（可以转移过去）

与长度无关，因此可以矩阵加速

但最后第一行的哪些答案是符合条件的，就还需要一次dfs，确定对于每一个$i$枚举的长度都必须至少是一个限制的倍数

到最后一层的时候，就去调用哈希字符串对应的编号，在矩阵中进行查询，累加到答案上

code

#include <map>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
#define mod 12345
#define int long long
vector < int > G[27];
int n, m, cnt, ret;
int lim[27];struct matrix {int c[125][125];matrix() { memset( c, 0, sizeof( c ) ); }int * operator [] ( int i ) { return c[i]; }matrix operator * ( matrix &t ) const {matrix ans;for( int i = 1;i <= 123;i ++ )for( int j = 1;j <= 123;j ++ )for( int k = 1;k <= 123;k ++ )ans[i][j] = ( ans[i][j] + c[i][k] * t[k][j] ) % mod;return ans;}
}g;matrix qkpow( matrix x, int y ) {matrix ans;for( int i = 1;i <= 123;i ++ )ans[i][i] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans;
}struct node {int cnt[27];node() { memset( cnt, 0, sizeof( cnt ) ); }bool operator < ( const node &t ) const {for( int i = 1;i <= 26;i ++ )if( cnt[i] == t.cnt[i] ) continue;else return cnt[i] < t.cnt[i];return 0;}
}h;map < node, int > mp;void dfs( int x, node s ) {if( x > 26 ) {mp[s] = ++ cnt;return;}if( ! lim[x] ) dfs( x + 1, s );for( int i = 0;i < lim[x];i ++ ) {s.cnt[x] = i;dfs( x + 1, s );}
}void calc( int x, node s ) {if( x > 26 ) {ret = ( ret + g[1][mp[s]] ) % mod;return;}if( ! lim[x] ) calc( x + 1, s );for( int i = 0;i < lim[x];i ++ ) {for( auto j : G[x] )if( i % j == 0 ) goto next;continue;next :s.cnt[x] = i;calc( x + 1, s );}
}signed main() {scanf( "%lld %lld", &n, &m );if( ! n ) return ! printf( "1\n" );if( ! m ) return ! printf( "0\n" );for( int i = 1;i <= m;i ++ ) {char ch; int x, c;scanf( "\n%c %lld", &ch, &x );c = ch - 'A' + 1;if( ! lim[c] ) lim[c] = x;else lim[c] *= x;G[c].push_back( x ); }dfs( 1, h );for( map < node, int > :: iterator it = mp.begin();it != mp.end();it ++ ) {int id = it -> second; node now = it -> first;for( int i = 1;i <= 26;i ++ ) {if( ! lim[i] ) continue;node t = now;t.cnt[i] = ( now.cnt[i] + 1 ) % lim[i];g[id][mp[t]] ++;}}g = qkpow( g, n );calc( 1, h );printf( "%lld\n", ret );return 0;
}

UVA12183 Top Secret

problem

$N$个数排成一圈，一次操作为，每个位置的数$+=L\times$左$+R\times$右，保留$x$为整数

问$S$轮操作后每个位置的值

$N<=1000,S<=2^{30},x<=9$​

solution

非常标准的矩乘转移加速题干

但是发现$n$​级别在$1e3$​，矩乘是$n^3$的复杂度，也就是说常规的矩乘不足以通过这道题

发现转移矩阵的每一行相当于上一行进行右移$1$位，这种特殊的矩阵称之为循环矩阵

只需要计算第一行的加速矩阵，然后每一行都可以根据一定的公示在第一行进行查找

code

#include <cstdio>
#include <cstring>
#define maxn 1005
#define int long long
int n, s, l, r, x, mod, T;
int ret[maxn], h[maxn];struct matrix {int n;int c[2][maxn];void clear() { memset( c, 0, sizeof( c ) ); }matrix() { clear(); }int * operator [] ( int i ) { return c[i]; }matrix operator * ( matrix &t ) const {matrix ans; ans.n = n;for( int k = 1;k <= n;k ++ )for( int j = 1;j <= n;j ++ ) {int i = ( j - k + 1 + n ) % n;if( ! i ) i = n;ans[1][j] = ( ans[1][j] + c[1][k] * t[1][i] ) % mod;}return ans;}
}g;matrix qkpow( matrix x, int y ) {matrix ans; ans.n = x.n; ans[1][1] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans;
}signed main() {scanf( "%lld", &T );while( T -- ) {scanf( "%lld %lld %lld %lld %lld", &n, &s, &l, &r, &x );mod = 1;for( int i = 1;i <= x;i ++ )mod = ( mod << 3 ) + ( mod << 1 );g.clear(); g.n = n;for( int i = 1;i <= n;i ++ )scanf( "%lld", &h[i] ), ret[i] = 0;g[1][1] = 1, g[1][n] = l, g[1][2] = r;g = qkpow( g, s );for( int k = 1;k <= n;k ++ )for( int i = 1;i <= n;i ++ ) {int j = ( k - i + 1 + n ) % n;if( ! j ) j = n;ret[i] = ( ret[i] + g[1][j] * h[k] ) % mod;}for( int i = 1;i < n;i ++ )printf( "%lld ", ret[i] );printf( "%lld\n", ret[n] );}return 0;
}

P3746 [六省联考2017]组合数问题

problem

solution

设$f_{i,j}:$ 前$i$个物品选择物品数量取模$k$为$j$的方案数

$f_{i,j}=f_{i-1,j}+f_{i-1,(j-1+k)\%k}$

$i$的转移只跟$i-1$有关，且$j$的转移也是固定的

可以矩乘加速

code

#include <cstdio>
#include <cstring>
using namespace std;
#define int long long
#define maxn 50
int n, p, K, r;
struct matrix {int c[maxn][maxn];matrix() {memset( c, 0, sizeof( c ) );}int * operator [] ( int i ) {return c[i];}matrix operator * ( matrix &t ) const {matrix ans;for( int i = 0;i < K;i ++ )for( int j = 0;j < K;j ++ )for( int k = 0;k < K;k ++ )ans[i][j] = ( ans[i][j] + c[i][k] * t[k][j] ) % p;return ans;} 
}ret, g;matrix qkpow( matrix x, int y ) {matrix ans;for( int i = 0;i < K;i ++ )ans[i][i] = 1;while( y ) {if( y & 1 ) ans = ans * x;x = x * x;y >>= 1;}return ans;
}signed main() {scanf( "%lld %lld %lld %lld", &n, &p, &K, &r );ret[0][0] = 1;for( int i = 0;i < K;i ++ )g[i][i] ++, g[i][( i - 1 + K ) % K] ++;g = qkpow( g, n * K );ret = ret * g;printf( "%lld\n", ret[0][r] );return 0;
}