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hdu 1757 A Simple Math Problem 矩阵
A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2831 Accepted Submission(s): 1693
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
#include<iostream> #include<cstring> using namespace std; typedef long long ll; struct matrix { ll f[11][11]; }; int n,m; matrix mul(matrix a,matrix b){ matrix s; memset(s.f,0,sizeof s.f); int i,j,k; for(k=0;k<10;k++) for(i=0;i<10;i++){ if(!a.f[i][k]) continue; for(j=0;j<10;j++){ if(!b.f[k][j]) continue; s.f[i][j]+=a.f[i][k]*b.f[k][j]; s.f[i][j]%=m; } } return s; } matrix pow_mod(matrix a,int k){ matrix s; memset(s.f,0,sizeof s.f); for(int i=0;i<10;i++) s.f[i][i]=1; while(k){ if(k&1) s=mul(s,a); a=mul(a,a); k>>=1; } return s; } int main(){ while(cin>>n>>m){ int i; matrix e; memset(e.f,0,sizeof e.f); for(i=0;i<10;i++) cin>>e.f[i][0]; if(n<10){ cout<<n<<endl; continue; } for(i=1;i<10;i++) e.f[i-1][i]=1; e=pow_mod(e,n-9); ll ans=0; for(i=0;i<10;i++) ans=(ans+(9-i)*e.f[i][0])%m; cout<<ans<<endl; } return 0; }
hdu 1757 A Simple Math Problem 矩阵
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