LightOJ 1006 - Hex-a-bonacci

1006 - Hex-a-bonacci

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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

 

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PROBLEM SETTER: JANE ALAM JAN

很明显不能直接按题上的代码直接敲，题意代码就是一个递归的过程，把这个过程模拟出来即可，（记得当时做时，一直想着有什么规律。。。。）

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int g[11000];int main(){int n,i,j,k,l,t,temp;scanf("%d",&t);k=1;while(t--){   memset(g,0,sizeof(g));       scanf("%d%d%d%d%d%d%d",&g[0],&g[1],&g[2],&g[3],&g[4],&g[5],&n);       for(i=6;i<=n;i++)       {         temp=0;         for(j=i-1;j>=i-6;j--)      {       temp+=g[j]%10000007;   }   g[i]=temp;   }   printf("Case %d: %d\n",k++,g[n]%10000007);}return 0; }