HDOJ 1757 A Simple Math Problem （矩阵快速幂）

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3695    Accepted Submission(s): 2214

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 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

 

Sample Output

45104

 



题意：题目不长，应该能看懂


思路：




ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 101000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)using namespace std;int ans[10][10];int a[10][10];int MOD;void Matrix(int A[10][10],int B[10][10],int n){int T[10][10];mem(T);for(int i=0;i<n;i++){for(int j=0;j<n;j++){for(int k=0;k<n;k++){T[i][j]+=A[i][k]*B[k][j];}T[i][j]=T[i][j]%MOD;}}for(int i=0;i<n;i++){for(int j=0;j<n;j++){A[i][j]=T[i][j];}}}void Matrixpow(int A[10][10],int n,int m){mem(ans);for(int i=0;i<n;i++)ans[i][i]=1;while(m){if(m%2)Matrix(ans,A,n);Matrix(A,A,n);m/=2;}}int main(){LL k;int num,i,j;while(scanf("%lld%d",&k,&MOD)!=EOF){mem(a);for(i=0;i<10;i++){scanf("%d",&num);a[0][i]=num;a[i+1][i]=1;}if(k<10)//特判{printf("%d\n",k%MOD);continue;}Matrixpow(a,10,k-9);int sum=0;for(i=0;i<10;i++){sum+=ans[0][i]*(9-i);sum=sum%MOD;}printf("%d\n",sum);}return 0;}