hdu 5015 233 Matrix(最快的搞法)

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 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 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, could you tell me an,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 ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 31).
Output For each case, output an,m mod 10000007.
Sample Input

1 1 1 2 2 0 0 3 7 23 47 16

Sample Output

234 2799 72937 Hint hdu 5015 233 Matrix(最快的搞法)

Source 2014 ACM/ICPC Asia Regional Xi"> Recommend hujie | We have carefully selected several similar problems for you: 5017 5016 5014 5013 5012

解释在代码里:

#include #define MOD 10000007 using namespace std; typedef long long LL; #define rep(i,a,b) for(int i=a;i>=1; } return result; } LL f(LL N) { if(N==0) return (tmp-1+MOD)*inv[9]%MOD;///用到了等比数列求和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;///第一行的和乘组合数 int i; for(i=1;i

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