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.

Sample Input

5
0 1 2 3 4 5 20
3 2 1 5 0 1 9
4 12 9 4 5 6 15
9 8 7 6 5 4 3
3 4 3 2 54 5 4

Output for Sample Input

Case 1: 216339
Case 2: 79
Case 3: 16636
Case 4: 6
Case 5: 54

题解：

很明显，递推。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod = 10000007;
int a, b, c, d, e, f;
int num[10050];
int main()
{
    int n, caseno = 0, cases;
    scanf("%d", &cases);
    while (cases--)
    {
        memset(num, 0, sizeof(num));
        scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);
        num[0] = a, num[1] = b, num[2] = c, num[3] = d, num[4] = e, num[5] = f;
        for (int i = 6; i <= n; i++)
        {
            for (int j = 1; j <= 6; j++)
                num[i] += num[i-j], num[i] %= mod;
        }
        printf("Case %d: %d\n", ++caseno, num[n] % 10000007);
    }
    return 0;
}