T1：Sequence

title

传送

solution

一眼就是很裸的矩阵加速
$⌊pl⌋\lfloor\frac{p}{l}\rfloor$ 分块矩阵加速就可以了
$[BA1]×[DC⌊pl⌋010001]\begin{bmatrix} B\\ A\\ 1\\ \end{bmatrix} \times \begin{bmatrix} D&C&\lfloor\frac{p}{l}\rfloor\\ 0&1&0\\ 0&0&1 \end{bmatrix}$
这道题唯一算得上是坑的应该是 $n, p$ 的大小，当 $p / l = = 0$ 直接矩阵加速到底即可，注意 $r$ 不能超过 $n$
在这里插入图片描述 ## code

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define ll long long
#define mod 1000000007
struct Matrix {ll c[5][5];void init() {memset( c, 0, sizeof( c ) );}Matrix operator * ( const Matrix &a ) {Matrix ans;ans.init();for( int i = 1;i <= 3;i ++ )for( int j = 1;j <= 3;j ++ )for( int k = 1; k <= 3;k ++ )ans.c[i][j] = ( ans.c[i][j] + c[i][k] * a.c[k][j] ) % mod;return ans;}
}V;
ll T, A, B, C, D, P, n;
ll ans1, ans2;Matrix qkpow( Matrix a, int b ) {Matrix ans;ans.init();for( int i = 1;i <= 3;i ++ ) ans.c[i][i] = 1;while( b ) {if( b & 1 ) ans = ans * a;a = a * a;b >>= 1;}return ans;
}signed main() {scanf( "%lld", &T );while( T -- ) {scanf( "%lld %lld %lld %lld %lld %lld", &A, &B, &C, &D, &P, &n );if( n == 1 ) { printf( "%lld\n", A ); continue; }if( n == 2 ) { printf( "%lld\n", B ); continue; }ans1 = B, ans2 = A;for( int l = 3, r;l <= n;l = r + 1 ) {if( P / l == 0 ) {V.init();V.c[1][1] = D, V.c[1][2] = C;V.c[2][1] = V.c[3][3] = 1;V = qkpow( V, n - l + 1 );ans1 = ( V.c[1][1] * ans1 % mod + V.c[1][2] * ans2 % mod + V.c[1][3] ) % mod;ans2 = ( V.c[2][1] * ans1 % mod + V.c[2][2] * ans2 % mod + V.c[2][3] ) % mod;break;}r = min( n, P / ( P / l ) );V.init();V.c[1][1] = D, V.c[1][2] = C, V.c[1][3] = P / l;V.c[2][1] = V.c[3][3] = 1;V = qkpow( V, r - l + 1 );ll newans1 = ( V.c[1][1] * ans1 % mod + V.c[1][2] * ans2 % mod + V.c[1][3] ) % mod;ll newans2 = ( V.c[2][1] * ans1 % mod + V.c[2][2] * ans2 % mod + V.c[2][3] ) % mod;ans1 = newans1, ans2 = newans2;}printf( "%lld\n", ans1 );} return 0;
}

T2：Bamboo Partition

title

传送门

solution

$_{i=1}^nd−((a_i−1)\%d+1)≤k$
$=∑i=1nd−∑i=1n(ai−1−⌊ai−1d⌋∗d+1)≤k=\sum_{i=1}^nd-\sum_{i=1}^n(a_i-1-\lfloor\frac{a_i-1}{d}\rfloor*d+1)\le k$
$=n∗d−∑i=1nai+∑i=1n⌊ai−1d⌋∗d≤k=n*d-\sum_{i=1}^na_i+\sum_{i=1}^n\lfloor\frac{a_i-1}{d}\rfloor*d\le k$
$d(n+∑i=1n⌊ai−1d⌋∗d)≤k+∑i=1naid(n+\sum_{i=1}^n\lfloor\frac{a_i-1}{d}\rfloor*d)\le k+\sum_{i=1}^na_i$
$⌊ai−1d⌋\lfloor\frac{a_i-1}{d}\rfloor$ 有根号的取值，我们直接分块即可
在这里插入图片描述

code

#include <cstdio>
#include <iostream>
using namespace std;
#define N 105
#define int long long
int n, k, Max, ans;
int a[N];signed main() {scanf( "%lld %lld", &n, &k );for( int i = 1;i <= n;i ++ ) scanf( "%lld", &a[i] ), k += a[i], Max = max( Max, a[i] - 1 );for( int l = 1, r, sum;l <= Max;l = r + 1 ) {r = Max, sum = 0;for( int i = 1;i <= n;i ++ )if( a[i] - 1 >= l ) {sum += ( a[i] - 1 ) / l;r = min( r, ( a[i] - 1 ) / ( ( a[i] - 1 ) / l ) );}if( l <= k / ( sum + n ) ) ans = max( ans, min( k / ( sum + n ), r ) );}if( Max < k / n ) ans = max( ans, k / n );printf( "%lld", ans );return 0;
}