A simple math problem
Time Limit: 3000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 2780 accepted submission (s): 1649
Problem descriptionlele 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 cocould you help Lele to caculate F (k) % m.
Inputthe 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.
Outputfor each case, Output F (k) % m in one line.
Sample Input
10 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0
Sample output
45104
Illustration:
This figure comes from: http://www.cnblogs.com/wally/archive/2013/03/01/2938305.html
#include"iostream"#include"stdio.h"#include"string.h"#include"algorithm"#include"queue"#include"vector"using namespace std;#define N 10#define LL __int64int M,f[N];struct Mat{ LL mat[N][N];};Mat operator *(Mat a,Mat b){ int i,j,k; Mat c; for(i=0;i<N;i++) { for(j=0;j<N;j++) { c.mat[i][j]=0; for(k=0;k<N;k++) { c.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%M; } } } return c;}int fun(Mat &a,int k){ int i; Mat ans; memset(ans.mat,0,sizeof(ans.mat)); k-=9; for(i=0;i<N;i++) ans.mat[i][i]=1; for(i=1;i<N;i++) a.mat[i][i-1]=1; while(k) { if(k&1) ans=ans*a; k>>=1; a=a*a; } LL s=0; for(i=0;i<N;i++) { s+=ans.mat[0][i]*f[N-i-1]%M; s%=M; } return s;}int main(){ int k,i; Mat a; while(scanf("%d%d",&k,&M)!=-1) { memset(a.mat,0,sizeof(a.mat)); for(i=0;i<N;i++) { scanf("%I64d",&a.mat[0][i]); f[i]=i; } if(k<N) printf("%d\n",i%M); else printf("%d\n",fun(a,k)); } return 0;}
HDU 1757 a simple math problem (matrix fast power)