Hdu 5015 233 Matrix (the fastest way), hdu5015

Hdu 5015 233 Matrix (the fastest way), hdu5015

233 Matrix Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 191 Accepted Submission (s): 125


Problem DescriptionIn our daily life we often use 233 to express our feelings. actually, we may say 2333,233 33, or 233333... in the same meaning. and here is the question: Suppose we have a matrix called 233 matrix. in the first line, it wocould be 233,233 3, 23333... (it means a0, 1 = 233, a0, 2 = 2333, a0, 3 = 23333 ...) besides, in 233 matrix, we got ai, j = ai-1, j + ai, J-1 (I, j = 0 ). now you have known A1, 0, a2, 0,..., an, 0, cocould you tell me an, m in the 233 matrix?
InputThere are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n, m (n ≤ 10, m ≤109 ). the second line contains n integers, a1, 0, a2, 0 ,..., an, 0 (0 ≤ ai, 0 <231 ).
OutputFor each case, output an, m mod 10000007.
Sample Input

1 112 20 03 723 47 16

 
Sample Output

234279972937Hint
  

 
Source2014 ACM/ICPC Asia Regional Xi 'an Online
Recommendhujie | We have carefully selected several similar problems for you: 5017 5016 5014 5013

In the Code:

# Include <cstdio> # define MOD 10000007 using namespace std; typedef long LL; # define rep (I, a, B) for (int I = a; I <B; I ++) LL inv [11]; // reverse LL inv_fac [11]; // factorial inverse LL n, m, a [11], tmp; /** after simulation, it is found that ans = c (m, 0) * a [n-1] + c (m +) * a [N-2] +... c (m + n, n) * a [0] + c (n, 0) * (23333 ...) + c (n + 1, 1) * (2333 ...) + c (n + m-1) * 233; deformation is as follows: ans = c (m, 0) * a [n-1] + c (m +) * a [N-2] +... c (m + n, n) * a [0] + c (n, n) * (23333 ...) + c (n + 1, n) * (2333 ...) + c (n + m-2, n) * 233; ans = C (m, 0) * a [n-1] + c (m +) * a [N-2] +... c (m + n, n) * a [0] + {c (n, n) + c (n + 1, n) + c (n + m-2, n )} * 233 + 2100*{10 ^ (m-2) * c (n, n) + 10 ^ (m-3) * c (n + 1, n) + 10 ^ 0 * c (n + M-4, n)}; ans = c (m, 0) * a [n-1] + c (m +) * a [N-2] +... c (m + n, n) * a [0] + c (n + s-1, n) * 233 + 2100*{10 ^ (m-2) * c (n, n) + 10 ^ (m-3) * c (n + 1, n) + 10 ^ 0 * c (n + M-4, n)}; set f (n) = {10 ^ (m-2) * c (n, n) + 10 ^ (m-3) * c (n + 1, n) + 10 ^ 0 * c (n + M-4, n)}, then f (n) = (10 * f (n-1)-c (n + m-2, n)/9) % MOD; that is, f (n) = (10 * f (n-1)-c (n + m-2, n) * inv [9] % MOD; **/v Oid Init () // initialization Array {int I; inv [0] = inv [1] = 1; inv_fac [0] = inv_fac [1] = 1; for (I = 2; I <15; I ++) {inv [I] = (inv [MOD % I] * (MOD-MOD/I) % MOD; /// Recursive Inverse element inv_fac [I] = (inv_fac [I-1] * inv [I]) % MOD; /// returns the inverse element of the factorial} ll c (ll n, LL k) // calculates the number of combinations {if (k = 0) return 1; LL I, ans = inv_fac [k]; for (I = 0; I <k; I ++) ans = ans * (N-I) % MOD; return ans ;} LL quick_mod (LL x, ll n) // fast power modulo {LL temp = x % MOD, result = 1; while (N) {if (N & 1) result = result * temp % MOD; temp = Temp * temp % MOD; N >>= 1;} return result;} LL f (ll n) {if (N = 0) return (tmp-1 + MOD) * inv [9] % MOD; // sum of the used proportional Series sum [n] = a1 (1-q ^ n)/(1-q); return (10 * f (N-1) -C (N + m-2, N) * inv [9] % MOD;} void solve () {if (n = 0 & m = 0) {puts ("0"); return;} tmp = quick_mod (10 m-1); LL ans = (C (n + m-1, n) * 233 + f (n) * 2100) % MOD; // The sum of the first line and the number of multiplication groups int I; for (I = 1; I <= n; I ++) ans = (ans + C (n + m-1-i, n-I) * a [I-1]) % MOD; /// add the first column and the number of multiplication groups printf ("% I64d \ n", ans);} int ma In () {Init (); while (~ Scanf ("% I64d % I64d", & n, & m) {rep (I, 0, n) {scanf ("% I64d", & a [I]); a [I] % = MOD;} solve ();} return 0 ;}