HDU 1757 A Simple Math Problem
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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
题意:按f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10) (x>=10) ; f(x) = x(x<10)来计算f(x)%m的值。
分析:这题要用递推,并且k值很大,所以需要用矩阵快速幂。
构造的矩阵是:
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* |
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= |
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写个结构类型代表矩阵,以及矩阵的相乘的函数和矩阵快速幂的函数,注意一下初始化。
#include<stdio.h> #include<string.h> int n,k,m; struct matrix { int a[15][15]; int row,col; void init(int r,int c){ memset(a,0,sizeof(a)); row=r;col=c; } } big,f,u; matrix mul(matrix a,matrix b) { matrix c; c.init(a.row,b.col); for(int i=0; i<a.row; i++) for(int j=0; j<b.col; j++) { for(int k=0; k<a.col; k++) c.a[i][j]+=(a.a[i][k]*b.a[k][j])%m; c.a[i][j]%=m; } return c; } void init() { big.init(10,10); f.init(10,1); for(int i=1; i<10; i++) big.a[i][i-1]=1; for(int i=0; i<10; i++) f.a[i][0]=9-i; } matrix qpow(matrix a,int k) { matrix ans; ans.init(a.row,a.col); for(int i=0;i<a.row;i++) ans.a[i][i]=1; while(k) { if(k&1)ans=mul(ans,a); a=mul(a,a); k>>=1; } return ans; } int main() { init(); while(~scanf("%d%d",&k,&m)) { for(int i=0; i<10; i++) scanf("%d",&big.a[0][i]); u=mul(qpow(big,k-9),f); printf("%d\n",u.a[0][0]%m); } return 0; }
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