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 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 1 1 2 2 0 0 3 7 23 47 16

 

Sample Output

234 2799 72937

Hint

 

Source

2014 ACM/ICPC Asia Regional Xi'an Online

 

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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<=n;i++)
        ans=(ans+C(n+m-1-i,n-i)*a[i-1])%MOD;///加上第一列的和乘组合数
    printf("%I64d\n",ans);
}
int main()
{
    Init();
    while(~scanf("%I64d%I64d",&n,&m))
    {
        rep(i,0,n){
            scanf("%I64d",&a[i]);
            a[i]%=MOD;
        }
        solve();
    }
    return 0;
}