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Hex-a-bonacci

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Given a code (not optimized), and necessary inputs, you have to find the output of the code for the inputs. The code is as follows:

int a, b, c, d, e, f;
int fn( int n ) {
    if( n == 0 ) return a;
    if( n == 1 ) return b;
    if( n == 2 ) return c;
    if( n == 3 ) return d;
    if( n == 4 ) return e;
    if( n == 5 ) return f;
    return( fn(n-1) + fn(n-2) + fn(n-3) + fn(n-4) + fn(n-5) + fn(n-6) );
}
int main() {
    int n, caseno = 0, cases;
    scanf("%d", &cases);
    while( cases-- ) {
        scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);
        printf("Case %d: %d\n", ++caseno, fn(n) % 10000007);
    }
    return 0;
}

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains seven integers, a, b, c, d, e, f and n. All integers will be non-negative and 0 ≤ n ≤ 10000 and the each of the others will be fit into a 32-bit integer.

Output

For each case, print the output of the given code. The given code may have integer overflow problem in the compiler, so be careful.

AC代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define H 10000007
 4 int shu[10100], n;
 5 int fun()
 6 {
 7     for(int  i = 0; i < 6; i++)
 8         shu[i] %= H;
 9     if(n < 6)
10         return shu[n];
11     for(int  i = 6; i <= n; i++)
12         shu[i]=(shu[i-6]+shu[i-5]+shu[i-4]+shu[i-3]+shu[i-2]+shu[i-1])%H;
13     return shu[n];
14 }
15 int main()
16 {
17     int t;scanf("%d",&t);
18     for(int j = 1; j <= t; j++)
19     {
20         scanf("%d%d%d%d%d%d%d",&shu[0],&shu[1],&shu[2],&shu[3],&shu[4],&shu[5],&n);
21         int ans = fun();
22         printf("Case %d: %d\n", j, ans);
23     }
24     return 0;
25  }