HDU 1757 矩陣乘法

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 560    Accepted Submission(s): 316

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.

 

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.

 

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

/* 由| 1 1 | * | F1 | = | F2 | 可以得到這題斐波那契數列可以用到矩陣解決 | 1 0 | | F0 | | F1 | */ #include<stdio.h> //此題採用矩陣乘法來做，並且要用到快速冪 struct node { int arc[10][10]; }; node b,tp,res,init; int n,m; node juzhen(node a,node b) { node c; int i,j,k; for(i=0;i<10;i++) { for(k=0;k<10;k++) { c.arc[i][k]=0; for(j=0;j<10;j++) { c.arc[i][k]+=(a.arc[i][j]*b.arc[j][k])%m; c.arc[i][k]%=m; } } } return c; } void f(int temp) { tp=b; b=res; while(temp) { if(temp&1) //與1計算後可以得到二進制最後一位是否爲1，爲1就乘，不是就不乘 b=juzhen(b,tp); // 假設temp的二進制是10111 那麼b依次保存爲b的1次方，3 //方，7次方 ，7次方，23次方 tp=juzhen(tp,tp); //tp保存的是初始矩陣b的2次方，4次方，8次方。。。等等 temp=temp>>1; //循環移位，右移一次等於除以2 } } int main() { int i,j,sum; for(i=0;i<10;i++) for(j=0;j<10;j++) if(i-1==j) init.arc[i][j]=1; else init.arc[i][j]=0; while(scanf("%d%d",&n,&m)!=EOF) { b=init; // b是初始矩陣，第一行爲a0——a9 for(i=0;i<10;i++) scanf("%d",&b.arc[0][i]); for(i=0;i<10;i++) for(j=0;j<10;j++) if(i==j) res.arc[i][j]=1; else res.arc[i][j]=0; if(n>=10) { sum=0; f(n-9); for(i=0;i<10;i++) sum+=(b.arc[0][i]*(9-i))%m;//最後要算n小於10,fun(x)=x; printf("%d/n",sum%m); } else printf("%d/n",n%m); } return 0; }