hdoj5015 233 Matrix【矩阵快速幂】
233 MatrixTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1555 Accepted Submission(s): 918
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).
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
2014 ACM/ICPC Asia Regional Xi'an Online
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解题思路：因为将第一列设为 23 a1,0 a2,0 a3,0 an,0 3
初始化矩阵 10 0 0 1 求m次幂即可
10 1 0 1
10 1 1 1
0 0 0 1
#include
#include
#include
#include
#include
#define MOD 10000007
using namespace std;
long long num[15];
long long a[15][15];
long long ans[15][15];
void Mulit(long long A[15][15],long long B[15][15],long long n){
long long D[15][15]={0};
for(long long i=0;i>=1;
}
}
int main()
{
long long n,m,i,j;
while(scanf("%lld%lld",&n,&m)!=EOF){
for(i=1;i<=n;++i){
scanf("%lld",&num[i]);
}
num[0]=23;num[n+1]=3;
memset(a,0,sizeof(a));n+=2;
for(i=0;i