題干：

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 ai,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

結(jié)題報告：

? ? ? ? ?依舊是按照列，找一個轉(zhuǎn)移矩陣，然后做運算，最后乘上之前保存的數(shù)組，得到想要的結(jié)果

AC代碼：

#include<bits/stdc++.h>using namespace std; const int MAX = 20 ; const int mod = 10000007 ; struct Matrix {long long mat[MAX][MAX]; }; int n,m; long long b[MAX]; Matrix mul(Matrix a,Matrix b) {Matrix c;memset(c.mat,0,sizeof(c.mat));for(int i=0; i<=n+1; i++) {for(int j=0; j<=n+1; j++) {c.mat[i][j]=0;for(int k=0; k<=n+1; k++) {if(a.mat[i][k]&&b.mat[k][j]) {c.mat[i][j]+=a.mat[i][k]*b.mat[k][j];//矩陣乘法c.mat[i][j]%=mod;}}}}return c;//返回乘完了之后的矩陣 } Matrix q_pow(Matrix a, int k) {Matrix ans;memset(ans.mat,0,sizeof(ans.mat));for(int i=0;i<=n+1;i++)ans.mat[i][i]=1;while(k)//使用快速冪的思想進行矩陣的m次方相乘{(lán)if(k&1) {ans=mul(ans,a);}k>>=1;a=mul(a,a);}return ans; }int main() {int m;Matrix a;//轉(zhuǎn)移矩陣 while(~scanf("%d%d",&n,&m) ){//初始化第一列 //b數(shù)組就是第一列，最后與轉(zhuǎn)移矩陣的m次方相乘得到anmfor(int i=1; i<=n; i++) {scanf("%lld",&b[i]);}b[0]=23;b[n+1]=3;//求a這個轉(zhuǎn)移矩陣。先初始化！因為你比如a13 這個地方的值就沒有更新到，因為他是0，所以運算之后就可能改變值了，所以下一組輸入的時候a13這里就不是0了，所以這里一定要memset一下。 memset(a.mat,0,sizeof(a.mat));for(int i = 0; i<=n; i++) {a.mat[i][0]=10;a.mat[i][n+1]=1;}a.mat[n+1][n+1]=1;for(int i = 1; i<=n; i++) {for(int j = 1; j<=i; j++) {a.mat[i][j]=1;}}Matrix end = q_pow(a,m);long long ans=0;for(int i = 0;i<=n+1;i++) {ans+=(end.mat[n][i]*b[i])%mod,ans%=mod;}printf("%lld\n",ans);}return 0; }

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總結(jié)

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