Codeforces Round #291 (Div. 2) E. Darth Vader and Tree 矩阵快速幂

E. Darth Vader and Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

When Darth Vader gets bored, he sits down on the sofa, closes his eyes and thinks of an infinite rooted tree where each node has exactly n sons, at that for each node, the distance between it an its i-th left child equals to di. The Sith Lord loves counting the number of nodes in the tree that are at a distance at most x from the root. The distance is the sum of the lengths of edges on the path between nodes.

But he has got used to this activity and even grew bored of it. 'Why does he do that, then?' — you may ask. It's just that he feels superior knowing that only he can solve this problem.

Do you want to challenge Darth Vader himself? Count the required number of nodes. As the answer can be rather large, find it modulo109 + 7.

Input

The first line contains two space-separated integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 109) — the number of children of each node and the distance from the root within the range of which you need to count the nodes.

The next line contains n space-separated integers di (1 ≤ di ≤ 100) — the length of the edge that connects each node with its i-th child.

Output

Print a single number — the number of vertexes in the tree at distance from the root equal to at most x.

Sample test(s)

input

3 31 2 3

output

8

Note

Pictures to the sample (the yellow color marks the nodes the distance to which is at most three)

题意是求x分解成子树（每条边为权值为f(i)）的总个数，可以得到递推公式

dp[x] = dp[n-f(1)] + dp[n-f(2)]+....dp[n-f(i)] + 1;

可以发现f(i)最大为100，所以可 以得到公式dp[x] = sum(dp[x-i]) + 1( 1=< i <= 100),这样总的复杂度为101*101*101*log(n);就可得到解了！

ans为结果矩阵，one为乘数矩阵。

#define INF9000000000000000000#define EPS(double)1e-9#define mod1000000007#define PI3.14159265358979//*******************************************************************************/#endif#define N 101#define MOD 1000000007int n,x,pri[100050];long long matrix[N][N];void multimatrix(long long aa[][101],long long bb[][101],long long ans[][101],int len){    long long c[101][101];    FI(len){        FJ(len){            c[i][j] = 0;            FK(len){                c[i][j] += aa[i][k] * bb[k][j];                c[i][j] %= mod;            }        }    }    FI(len){        FJ(len){            ans[i][j] = c[i][j];        }    }}long long twoMulti(int x){    long long one[101][101],ans[101][101];    memset(one,0,sizeof(one));    memset(ans,0,sizeof(ans));    ans[0][0]  = ans[0][1]= 1;    FI(n)        one[pri[i]][1]++;    one[0][1] = one[0][0] = 1;    for(int i=1;i<100;i++){        one[i][i+1]  = 1;    }    while(x){        if(x&1){           multimatrix(ans ,one,ans,101);        }        multimatrix(one,one,one,101);        x/=2;    }    return ans[0][1];}int main(){    while (S2(n,x) != EOF)    {        FI(n){            S(pri[i]);        }        cout<<twoMulti(x)<<endl;    }    return 0;}