A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4331 Accepted Submission(s): 2603
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
题意:给出递推公式求第n项
题解:矩阵快速幂,注意初始矩阵为[9,8,7,6,5,4,3,2,1,0]
#include<bits/stdc++.h> #define N 10 #define mes(x) memset(x, 0, sizeof(x)); #define ll __int64 long long mod = 1e9+7; const int MAX = 0x7ffffff; using namespace std; struct mat{ ll a[N][N]; mat(){ memset(a, 0, sizeof(a)); } mat operator *(mat b){ mat c; for(int i=0;i<N;i++) for(int j=0;j<N;j++) for(int k=0;k<N;k++) c.a[i][j] = (c.a[i][j] + a[i][k]*b.a[k][j])%mod; return c; } }; mat f(mat x,ll m){ mat t; for(int i=0;i<N;i++) t.a[i][i] = 1; while(m){ if(m&1) t = t*x; x = x*x; m >>= 1; } return t; } int main() { ll k, i; while(~scanf("%I64d%I64d", &k, &mod)){ if(k<10){ printf("%I64d ", k); continue; } mat A, s; for(i=0;i<10;i++) s.a[0][i] = 9-i; for(i=0;i<10;i++) scanf("%I64d", &A.a[i][0]); for(i=0;i<9;i++) A.a[i][i+1] = 1; s = s*f(A, k-9); printf("%I64d ", s.a[0][0]); } return 0; }