233 Matrix
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4591 Accepted Submission(s): 2575
Problem Description
In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a
0,1 = 233,a
0,2 = 2333,a
0,3 = 23333...) Besides, in 233 matrix, we got a
i,j = a
i-1,j +a
i,j-1( i,j ≠ 0). Now you have known a
1,0,a
2,0,...,a
n,0, could you tell me a
n,m in the 233 matrix?
Input
There are multiple test cases. Please process till EOF.
For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).
For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).
Output
For each case, output a
n,m mod 10000007.
Sample Input
1 1 1 2 2 0 0 3 7 23 47 16
Sample Output
234 2799 72937
Hint
Source
Recommend
This question is still the main ideas: one a recursive past
In Case of n = 3 the matrix
a1,1 1 0 0 1 0 a1,0
a2,1 1 1 0 1 0 a2,0
a3,1 1 1 1 1 0 a3,0
2333 0 0 0 10 3 233
1 0 0 0 0 1 1
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <algorithm> #include <iostream> #include<cstdio> #include<string> #include<cstring> #include <stdio.h> #include <string.h> #include <vector> #define ME(x , y) memset(x , y , sizeof(x)) #define SF(n) scanf("%d" , &n) #define rep(i , n) for(int i = 0 ; i < n ; i ++) #define INF 0x3f3f3f3f #define mod 10000007 using namespace std; typedef long long ll ; ll n , m ; struct node{ ll a[19][19]; node() { memset(a , 0 , sizeof(a)); } }; node mul(node A , node B) { node C ; for(int i = 0 ; i < n + 2 ; i++) { for(int j = 0 ; j < n + 2 ; j++) { for(int k = 0 ; k < n + 2 ; k++) { C.a[i][j] = (C.a[i][j] + A.a[i][k]*B.a[k][j]) % mod; } } } return C ; } node pow1(node A , ll t) { node ans ; for(int i = 0 ; i < n + 2; i++) ans.a[i][i] = 1 ; while(t) { if(t&1) { ans = mul(ans , A); } t >>= 1 ; A = mul(A , A); } return ans ; } int main() { while(~scanf("%d%d" , &n , &m)) { node A , B , C ; for(int i = 0 ; i < n ; i++) { for(int j = 0 ; j < n + 2 ; j++) { if(j == n) A.a[i][j] = 1 ; if(j <= i) A.a[i][j] = 1 ; } } A.a[n][n] = 10 , A.a[n][n + 1] = 3 ; A.a[n + 1][n + 1] = 1 ; for(int i = 0 ; i < n ; i++) { for(int j = 0 ; j < n ; j++) { if(j == 0) scanf("%lld" , &B.a[i][j]); } } B.a[n][0] = 233 ; B.a[n+1][0] = 1 ; C = mul(pow1(A , m) , B); printf("%lld\n" , C.a[n - 1][0]); } return 0 ; }