light OJ 1006 solution in c++

light OJ 1006 problem link: http://lightoj.com/volume_showproblem.php?problem=1006

light OJ 1006 problem: 1006 - Hex-a-bonacci

Time Limit: 0.5 second(s)

Memory Limit: 32 MB

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

my solution in c++ to light OJ 1006 Hex-a-bonacci problem link: http://ideone.com/pYjAzX

my solution in c++ to light OJ 1006 Hex-a-bonacci problem: 

#include <iostream>

#include <cstring>

#include <cstdio>

#define maxn 10003

using namespace std;

long long fn[maxn];

int main() {

// your code goes here

int n, cases, caseno=0, i;

cin >> cases;

while (cases--) {

fn[6]=0;

for (i=0;i<6;i++) {

scanf("%lld",&fn[i]); fn[i]=fn[i]%10000007;

fn[6]=(fn[6]+fn[i])%10000007;

}

cin >> n;

for (i=7;i<=n;i++) {

fn[i]=(fn[i-6]+fn[i-5]+fn[i-4]+fn[i-3]+fn[i-2]+fn[i-1])%10000007;

}

cout<<"Case "<<++caseno<<": "<<fn[n]<<endl;

}

return 0;

}

and here is my friend Quan's solution in c: http://ideone.com/HQvLu3

#include <cstdio>

#include <cstring>

#define maxn 10003

using namespace std;

long long fn[maxn];

int main() {

    int n, caseno = 0, cases,i;

    scanf("%d", &cases);

    while( cases-- ) {

    fn[6]=0;

    for (i=0;i<6;i++){

    scanf("%lld",&fn[i]); fn[i] = fn[i] % 10000007;

    fn[6]=(fn[6]+fn[i]) % 10000007;

    }

    scanf("%d",&n);

    for (i=7;i<=n;i++) fn[i]=(fn[i-1]+fn[i-2]+fn[i-3]+fn[i-4]+fn[i-5]+fn[i-6])% 10000007;

    //fn[i]=((2*fn[i-1]*10000007)-fn[i-7]) % 10000007;

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

    }

    return 0;

}