水题一枚

Description

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

 

Sample Output

Case 1: 216339

Case 2: 79

Case 3: 16636

Case 4: 6

Case 5: 54

#include <stdio.h>

#include <string.h>

int a, b, c, d, e, f;

int m[10005];



int main() {

    int n, caseno = 0, cases;

    scanf("%d", &cases);

    int flag=1;

    while(cases--){

        scanf("%d %d %d %d %d %d %d", &a, &b, &c, &d, &e, &f, &n);

        m[0]=a;

        m[1]=b;

        m[2]=c;

        m[3]=d;

        m[4]=e;

        m[5]=f;

        for(int i=6;i<=n;i++)

            m[i]=m[i-1]% 10000007+m[i-2]% 10000007+m[i-3]% 10000007+m[i-4]% 10000007+m[i-5]% 10000007+m[i-6]% 10000007;

        printf("Case %d: %d\n", ++caseno,m[n]% 10000007);

    }

    return 0;

}