LightOJ 1006 :Hex-a-bonacci

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 ", ++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

打表缩短时间

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int fn[100001];
int a, b, c, d, e, f;
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);
    fn[0]=a;
    fn[1]=b;
    fn[2]= c;
     fn[3]= d;
    fn[4]=e;
    fn[5]= f;
   for(int i=6;i<=n;i++)
  {
    fn[i]=fn[i-1]% 10000007 + fn[i-2]% 10000007 + fn[i-3]% 10000007 + fn[i-4]% 10000007 + fn[i-5] % 10000007+ fn[i-6]% 10000007 ;
   }
        printf("Case %d: %d
", ++caseno, fn[n] % 10000007);
    }
    return 0;
}