杭电ACM-LCY算法进阶培训班-专题训练（矩阵快速幂）【模板】

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杭电ACM-LCY算法进阶培训班-专题训练（矩阵快速幂）【模板】
- 矩阵快速幂模板
- Count
- - Problem Description
  - Input
  - Output
  - Sample Input
  - Sample Output
  - Source
  - 思路
- Sum of Tribonacci Numbers
- - Problem Description
  - Input
  - Output
  - Sample Input
  - Sample Output
  - Author
  - Source
  - 思路
- Another kind of FibonacciProblem Description
- - Input
  - Output
  - Sample Input
  - Sample Output
  - Author
  - Source
  - 思路
- Kiki & Little Kiki 2
- - Problem Description
  - Input
  - Output
  - Sample Input
  - Sample Output
  - Source
  - 思路

矩阵快速幂模板

void mul(){int c[maxn];memset(c,0,sizeof(c));for(int j=0;j<len;j++)for(int k=0;k<len;k++)c[j]=(c[j]+a[j][k]*b[k])%mod;memcpy(b,c,sizeof(b));
}void mulSelf(){int c[maxn][maxn];memset(c,0,sizeof(c));for(int i=0;i<len;i++)for(int j=0;j<len;j++)for(int k=0;k<len;k++)c[i][j]=(c[i][j]+a[i][k]*a[k][j])%mod;memcpy(a,c,sizeof(c));
}int main(){for(;n;n>>=1){if(n&1) mul();mulSelf();}
}

Count

Problem Description

Farmer John有n头奶牛.
某天奶牛想要数一数有多少头奶牛,以一种特殊的方式:
第一头奶牛为1号,第二头奶牛为2号,第三头奶牛之后,假如当前奶牛是第n头,那么他的编号就是2倍的第n-2头奶牛的编号加上第n-1头奶牛的编号再加上自己当前的n的三次方为自己的编号.
现在Farmer John想知道,第n头奶牛的编号是多少,估计答案会很大,你只要输出答案对于123456789取模.

Input

第一行输入一个T,表示有T组样例
接下来T行,每行有一个正整数n,表示有n头奶牛 (n>=3)
其中，T= 1 0 4 10^4 104, n < = 1 0 18 n<=10^{18} n<=1018

Output

共T行,每行一个正整数表示所求的答案

Sample Input

Sample Output

Source

“字节跳动-文远知行杯”广东工业大学第十四届程序设计竞赛

思路

递推公式为： d p [ i ] = d p [ i − 1 ] + 2 d p [ i − 2 ] + n 3 dp[i]=dp[i-1]+2dp[i-2]+n^3 dp[i]=dp[i−1]+2dp[i−2]+n3。但是这种形式无法直接用矩阵快速幂来解。
因为： n 3 = ( n − 1 + 1 ) 3 = ∑ k = 0 3 C 3 k ( n − 1 ) k = 1 + 3 ( n − 1 ) + 3 ( n − 1 ) 2 + ( n − 1 ) 3 n^3=(n-1+1)^3=\sum_{k=0}^3C_3^k(n-1)^k=1+3(n-1)+3(n-1)^2+(n-1)^3 n3=(n−1+1)3=∑k=03C3k(n−1)k=1+3(n−1)+3(n−1)2+(n−1)3，可以构造下面的矩阵：
[ f n f n − 1 n 3 n 2 n 1 ] = [ 1 2 1 3 3 1 1 0 0 0 0 0 0 0 1 3 3 1 0 0 0 1 2 1 0 0 0 0 1 1 0 0 0 0 0 1 ] × [ f n − 1 f n − 2 ( n − 1 ) 3 ( n − 1 ) 2 n − 1 1 ] \begin{bmatrix} f_n\\ f_{n-1}\\ n^3\\ n^2\\ n\\ 1\\ \end{bmatrix}= \begin{bmatrix} 1&2&1&3&3&1\\ 1&0&0&0&0&0\\ 0&0&1&3&3&1\\ 0&0&0&1&2&1\\ 0&0&0&0&1&1\\ 0&0&0&0&0&1\\ \end{bmatrix}\times \begin{bmatrix} f_{n-1}\\ f_{n-2}\\ (n-1)^3\\ (n-1)^2\\ n-1\\ 1\\ \end{bmatrix} ⎣⎢⎢⎢⎢⎢⎢⎡fnfn−1n3n2n1⎦⎥⎥⎥⎥⎥⎥⎤=⎣⎢⎢⎢⎢⎢⎢⎡110000200000101000303100303210101111⎦⎥⎥⎥⎥⎥⎥⎤×⎣⎢⎢⎢⎢⎢⎢⎡fn−1fn−2(n−1)3(n−1)2n−11⎦⎥⎥⎥⎥⎥⎥⎤

#include<iostream>
#include<cstring>
#define int long long
using namespace std;
const int mod=123456789;
int bCP[6][6]={{1,2,1,3,3,1},{1,0,0,0,0,0},{0,0,1,3,3,1},{0,0,0,1,2,1},{0,0,0,0,1,1},{0,0,0,0,0,1}},b[6][6];
int aCP[6]={2,1,8,4,2,1},a[6];void mul(){int c[6];memset(c,0,sizeof(c));for(int j=0;j<6;j++)for(int k=0;k<6;k++)c[j]=(c[j]+a[k]*b[j][k])%mod;memcpy(a,c,sizeof(a));
}void mulSelf(){int c[6][6];memset(c,0,sizeof(c));for(int i=0;i<6;i++)for(int j=0;j<6;j++)for(int k=0;k<6;k++)c[i][j]=(c[i][j]+b[i][k]*b[k][j])%mod;memcpy(b,c,sizeof(b));
}void solve(){int n;memcpy(a,aCP,sizeof(a));memcpy(b,bCP,sizeof(b));cin>>n; n-=2;for(;n;n>>=1){if(n&1) mul();mulSelf();}cout<<a[0]<<"\n";
}signed main(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int te;cin>>te;while(te--) solve();
}

Sum of Tribonacci Numbers

Problem Description

Everybody knows Fibonacci numbers, now we are talking about the Tribonacci numbers:
T[0] = T[1] = T[2] = 1;
T[n] = T[n - 1] + T[n - 2] + T[n - 3] (n >= 3)

Given a and b, you are asked to calculate the sum from the ath Fibonacci number to the bth Fibonacci number, mod 1,000,000,007, that is (T[a] + T[a + 1] + … + T[b]) % 1,000,000,007.

Input

There are multiple cases (no more than 100).

Each case contain two non-negative integers a b(a <= b and a, b < 1,000,000,000)

Output

For each case, output the sum % 1,000,000,007.

Sample Input

0 2
0 5

Sample Output

3
20

Author

hhanger@zju

Source

HDOJ 2009 Summer Exercise（5）

思路

设S[i]为数列T[i]的前缀和，那么易得 S [ i ] = S [ i − 1 ] + T [ i ] S[i]=S[i-1]+T[i] S[i]=S[i−1]+T[i]。可以构造如下矩阵：

[ S i T i T i − 1 T i − 1 ] = [ 1 1 1 1 0 1 1 1 0 1 0 0 0 0 1 0 ] × [ S i − 1 T i − 1 T i − 2 T i − 3 ] \begin{bmatrix} S_i\\ T_i\\ T_{i-1}\\ T_{i-1}\\ \end{bmatrix}= \begin{bmatrix} 1&1&1&1\\ 0&1&1&1\\ 0&1&0&0\\ 0&0&1&0\\ \end{bmatrix} \times \begin{bmatrix} S_{i-1}\\ T_{i-1}\\ T_{i-2}\\ T_{i-3}\\ \end{bmatrix} ⎣⎢⎢⎡SiTiTi−1Ti−1⎦⎥⎥⎤=⎣⎢⎢⎡1000111011011100⎦⎥⎥⎤×⎣⎢⎢⎡Si−1Ti−1Ti−2Ti−3⎦⎥⎥⎤

#include<iostream>
#include<cstring>
#define int long long
using namespace std;
const int mod=1000000007;
int aCP[4][4]={{1,1,1,1},{0,1,1,1},{0,1,0,0},{0,0,1,0}},bCP[4]={3,1,1,1},a[4][4],b[4];void mul(){int c[4];memset(c,0,sizeof(c));for(int j=0;j<4;j++)for(int k=0;k<4;k++)c[j]=(c[j]+a[j][k]*b[k])%mod;memcpy(b,c,sizeof(b));
}void mulSelf(){int c[4][4];memset(c,0,sizeof(c));for(int i=0;i<4;i++)for(int j=0;j<4;j++)for(int k=0;k<4;k++)c[i][j]=(c[i][j]+a[i][k]*a[k][j])%mod;memcpy(a,c,sizeof(a));
}int calc(int n){memcpy(a,aCP,sizeof(a));memcpy(b,bCP,sizeof(b));if(n<=2)    return n+1;n-=2;for(;n;n>>=1){if(n&1) mul();mulSelf();}return b[0];
}signed main(){ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);int x,y;while(cin>>x>>y){cout<<(calc(y)-calc(x-1)+mod)%mod<<'\n';    //防止结果为负}
}

Another kind of FibonacciProblem Description

As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , $S(N) = A ( 0 ) 2 + A ( 1 ) 2 + … … + A ( n ) 2 A(0)^2 +A(1)^2+……+A(n)^2 A(0)2+A(1)2+……+A(n)2.

Input

There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 2 31 2^{31} 231 – 1
X : 2<= X <= 2 31 2^{31} 231– 1
Y : 2<= Y <= 2 31 2^{31} 231 – 1

Output

For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.

Sample Input

2 1 1
3 2 3

Sample Output

6
196

Author

wyb

Source

HDOJ Monthly Contest – 2010.02.06

思路

设S[n]为A[n]平方的前缀和，显然： S [ i ] = S [ i − 1 ] + A [ i ] 2 S[i]=S[i-1]+A[i]^2 S[i]=S[i−1]+A[i]2
又 A [ i ] = X A [ i − 1 ] + Y A [ i − 2 ] A[i]=XA[i-1]+YA[i-2] A[i]=XA[i−1]+YA[i−2]，那么：
A [ i ] 2 = X 2 A [ i − 1 ] 2 + 2 X Y A [ i − 1 ] A [ i − 2 ] + Y 2 A [ i − 2 ] 2 A[i]^2=X^2A[i-1]^2+2XYA[i-1]A[i-2]+Y^2A[i-2]^2 A[i]2=X2A[i−1]2+2XYA[i−1]A[i−2]+Y2A[i−2]2
并且 A [ i ] A [ i − 1 ] = X A [ i − 1 ] 2 + Y A [ i − 1 ] A [ i − 2 ] A[i]A[i-1]=XA[i-1]^2+YA[i-1]A[i-2] A[i]A[i−1]=XA[i−1]2+YA[i−1]A[i−2]
所以可以构造矩阵：
[ S i A i 2 A i − 1 2 A i A i − 1 ] = [ 1 x 2 y 2 2 x y 0 x 2 y 2 2 x y 0 1 0 0 0 x 0 y ] × [ S i − 1 A i − 1 2 A i − 2 2 A i − 1 A i − 2 ] \begin{bmatrix} S_i\\ A^2_i\\ A^2_{i-1}\\ A_iA_{i-1}\\ \end{bmatrix}= \begin{bmatrix} 1&x^2&y^2&2xy\\ 0&x^2&y^2&2xy\\ 0&1&0&0\\ 0&x&0&y\\ \end{bmatrix} \times \begin{bmatrix} S_{i-1}\\ A^2_{i-1}\\ A^2_{i-2}\\ A_{i-1}A_{i-2}\\ \end{bmatrix} ⎣⎢⎢⎡SiAi2Ai−12AiAi−1⎦⎥⎥⎤=⎣⎢⎢⎡1000x2x21xy2y2002xy2xy0y⎦⎥⎥⎤×⎣⎢⎢⎡Si−1Ai−12Ai−22Ai−1Ai−2⎦⎥⎥⎤

#include<iostream>
#include<cstring>
#define int long long
using namespace std;
const int mod=10007;
int a[4][4],b[4];void mul(){int c[4];memset(c,0,sizeof(c));for(int j=0;j<4;j++)for(int k=0;k<4;k++)c[j]=(c[j]+a[j][k]*b[k])%mod;memcpy(b,c,sizeof(c));
}void mulSelf(){int c[4][4];memset(c,0,sizeof(c));for(int i=0;i<4;i++)for(int j=0;j<4;j++)for(int k=0;k<4;k++)c[i][j]=(c[i][j]+a[i][k]*a[k][j])%mod;memcpy(a,c,sizeof(a));
}signed main(){ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);int n,x,y;while(cin>>n>>x>>y){x%=mod;y%=mod;a[0][0]=1,a[0][1]=x*x%mod,a[0][2]=y*y%mod,a[0][3]=2*x*y%mod;a[1][0]=0,a[1][1]=x*x%mod,a[1][2]=y*y%mod,a[1][3]=2*x*y%mod;a[2][0]=0,a[2][1]=1,a[2][2]=0,a[2][3]=0;a[3][0]=0,a[3][1]=x%mod,a[3][2]=0,a[3][3]=y%mod;b[0]=2,b[1]=1,b[2]=1,b[3]=1;for(n-=1;n;n>>=1){if(n&1) mul();mulSelf();}cout<<b[0]<<"\n";}
}

Kiki & Little Kiki 2

Problem Description

There are n lights in a circle numbered from 1 to n. The left of light 1 is light n, and the left of light k (1< k<= n) is the light k-1.At time of 0, some of them turn on, and others turn off.
Change the state of light i (if it’s on, turn off it; if it is not on, turn on it) at t+1 second (t >= 0), if the left of light i is on !!! Given the initiation state, please find all lights’ state after M second. (2<= n <= 100, 1<= M<= 1 0 8 10^8 108)

Input

The input contains one or more data sets. The first line of each data set is an integer m indicate the time, the second line will be a string T, only contains ‘0’ and ‘1’ , and its length n will not exceed 100. It means all lights in the circle from 1 to n.
If the ith character of T is ‘1’, it means the light i is on, otherwise the light is off.

Output

For each data set, output all lights’ state at m seconds in one line. It only contains character ‘0’ and '1.

Sample Input

Sample Output

1111000
001000010

Source

HDU 8th Programming Contest Site（1）

思路

d p [ i ] [ j ] = { d p [ i − 1 ] [ j ] , d p [ i − 1 ] [ j − 1 ] = 0 d p [ i − 1 ] [ j ] ⊗ 1 , d p [ i − 1 ] [ j − 1 ] = 1 dp[i][j]=\left\{ \begin{aligned} dp[i-1][j] & ,dp[i-1][j-1]=0 \\ dp[i-1][j] \otimes1 & ,dp[i-1][j-1]=1 \end{aligned} \right. dp[i][j]={dp[i−1][j]dp[i−1][j]⊗1,dp[i−1][j−1]=0,dp[i−1][j−1]=1
当取mod=2时，上式化为：
d p [ i ] [ j ] = d p [ i − 1 ] [ j ] + d [ i − 1 ] [ j − 1 ] dp[i][j]=dp[i-1][j]+d[i-1][j-1] dp[i][j]=dp[i−1][j]+d[i−1][j−1]

#include<cstdio>
#include<cstring>
using namespace std;
const int mod=2,maxn=105;
int a[maxn][maxn],b[maxn],n,ch,len;void mul(){int c[maxn];memset(c,0,sizeof(c));for(int j=0;j<len;j++)for(int k=0;k<len;k++)c[j]=(c[j]+a[j][k]*b[k])%mod;memcpy(b,c,sizeof(b));
}void mulSelf(){int c[maxn][maxn];memset(c,0,sizeof(c));for(int i=0;i<len;i++)for(int j=0;j<len;j++)for(int k=0;k<len;k++)c[i][j]=(c[i][j]+a[i][k]*a[k][j])%mod;memcpy(a,c,sizeof(c));
}int main(){while(~scanf("%d",&n)){getchar();for(len=0;(ch=getchar())!='\n';len++)b[len]=ch-'0';for(int i=0;i<len;i++)a[i][i]=a[i][(i-1+len)%len]=1;for(;n;n>>=1){if(n&1) mul();mulSelf();}for(int i=0;i<len;i++)    putchar(b[i]+'0');puts("");memset(a,0,sizeof(a));}
}

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杭电ACM-LCY算法进阶培训班-专题训练（矩阵快速幂）

杭电ACM-LCY算法进阶培训班-专题训练（矩阵快速幂）【模板】

矩阵快速幂模板

Count

Problem Description

Input

Output

Sample Input

Sample Output

Source

思路

Sum of Tribonacci Numbers

Problem Description

Input

Output

Sample Input

Sample Output

Author

Source

思路

Another kind of FibonacciProblem Description

Input

Output

Sample Input

Sample Output

Author

Source

思路

Kiki & Little Kiki 2

Problem Description

Input

Output

Sample Input

Sample Output

Source

思路

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