233矩阵（上三角矩阵求累加量//代码和思路不一样

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

1 112 20 03 723 47 16

Sample Output

234279972937          

题意就是a _i,j = a_i-1,j +a _i,j-1;

下面是例子。

#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <cstdio>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>using namespace std;#define pi acos(-1)#define endl '\n'#define rand() srand(time(0));#define me(x) memset(x,0,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0);typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;//const int dx[]={-1,0,1,0,-1,-1,1,1};//const int dy[]={0,1,0,-1,1,-1,1,-1};const int maxn=1e3+5;const int maxx=1e5+100;const double EPS=1e-7;const int MOD=10000007;#define mod(x) ((x)%MOD);template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;/*lch[root] = build(L1,p-1,L2+1,L2+cnt);    rch[root] = build(p+1,R1,L2+cnt+1,R2);中前*//*lch[root] = build(L1,p-1,L2,L2+cnt-1);    rch[root] = build(p+1,R1,L2+cnt,R2-1);中后*/long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}struct node{    LL t[13][13];    void mex()    {        me(t);    }}a,b,c,d,e;int n,m;node operator*(node a,node b)//重载运算符{    node ret;    LL x;    ret.mex();    for(int i=0;i<n;i++)        for(int j=0;j<n;j++)    {        x=0;        for(int k=0;k<n;k++)            x+=mod((LL)a.t[i][k]*b.t[k][j]);        ret.t[i][j]=mod(x);    }    return ret;}void init(){    a.mex();b.mex();c.mex();d.mex();e.mex();}node pow_mat(node a,node b,int x)//矩阵快速幂{    node ret=b;    while(x)    {        if(x%2)  ret=ret*a;        a=a*a;        for(int i=0;i<n;i++)        {            /*for(int j=0;j<n;j++)            {                cout<<ret.t[i][j]<<" ";            }            cout<<endl;*/        }        x>>=1;    }    return ret;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        init();        for(int i=n-1; i>=0; i--)            cin>>b.t[0][i];        //cout<<"输入完成"<<endl;        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                if(i>=j)                    c.t[i][j]=1;//上三角矩阵行列式求累加量        b=pow_mat(c,b,m);       // cout<<"-----------------"<<endl;        n=n+2;        d.t[0][n-1]=3;        d.t[0][n-2]=23;//构造右边233的矩阵        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                if(i>=j)                {                    if(i!=n-2)                        e.t[i][j]=1;                    else                        e.t[i][j]=10;                }        /*cout<<"-------------"<<endl;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                cout<<d.t[i][j]<<" ";            }            cout<<endl;        }        cout<<"-----------------"<<endl;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                cout<<e.t[i][j]<<" ";            }            cout<<endl;        }*/        d=pow_mat(e,d,m);        cout<<(d.t[0][0]+b.t[0][0])%MOD<<endl;    }}