A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4038 Accepted Submission(s):
2437
Problem Description
Lele now is thinking about a simple function
f(x).
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
Input
The problem contains mutiple test cases.Please process
to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
Output
For each case, output f(k) % m in one line.
Sample Input
10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0
Sample Output
45
104
Author
linle
Source
Recommend
典型的矩阵快速幂问题，第一次写，参考了别人的代码，感觉收益良多，最重要的地方在于构建一个矩阵。
借用一下别人代码的推论：
下面我们研究一下这道题如何运用矩阵。
f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)
构造的矩阵是:
|0 1 0 ......... 0| |f0| |f1 |
|0 0 1 0 ....... 0| |f1| |f2 |
|................1| * |..| = |...|
|a9 a8 .........a0| |f9| |f10|
*/
我们看到规律了，每次要到下次个A*B，以此类推则由A*A*A.......A*B；
题意：按照题目给的公式，求f(k) % m的值。
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 struct mat 6 { 7 int m[10][10]; 8 }; 9 int mod; 10 11 mat mul(mat a,mat b) 12 { 13 mat c; 14 int i,j,k; 15 memset(c.m,0,sizeof(c.m)); 16 for(i=0; i<10; i++) 17 for(j=0; j<10; j++) 18 { 19 for(k=0; k<10; k++) 20 { 21 c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod; 22 } 23 c.m[i][j]%=mod; 24 } 25 return c; 26 } 27 28 mat product(mat a,int k) 29 { 30 if(k==1) return a; 31 else if(k&1) 32 return mul(product(a,k-1),a); 33 else 34 return product(mul(a,a),k/2); 35 } 36 37 int main() 38 { 39 int i,j,k; 40 mat a; 41 while(~scanf("%d%d",&k,&mod)) 42 { 43 for(i=9; i>=0; i--) 44 scanf("%d",&a.m[9][i]); 45 if(k<10) 46 { 47 printf("%d ",k%mod); 48 continue; 49 } 50 for(i=0; i<9; i++) 51 for(j=0; j<10; j++) 52 if(j==i+1) a.m[i][j]=1; 53 else a.m[i][j]=0; 54 mat b=product(a,k-9); 55 int ans=0; 56 for(i=0; i<10; i++) 57 ans+=(b.m[9][i]*i)%mod; 58 printf("%d ",ans%mod); 59 } 60 return 0; 61 }