lightoj 1006 Hex-a-bonacci【水】

fuck!!!!!该死的题，说好的一大串，生怕按题中代码执行会超时，md，居然就是按题中的代码来，就改一丁点就行了！！！！！！！fuck，坑啊！！！！！！！！！

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

 

AC代码：

#include<stdio.h>
#include<string.h>
#define H 10000007
int shu[10100], n;
int fun()
{
    for(int  i = 0; i < 6; i++)
        shu[i] %= H;
    if(n < 6)
        return shu[n];
    for(int  i = 6; i <= n; i++)
        shu[i]=(shu[i-6]+shu[i-5]+shu[i-4]+shu[i-3]+shu[i-2]+shu[i-1])%H;
    return shu[n];
}
int main()
{
    int t;scanf("%d",&t);
    for(int j = 1; j <= t; j++)
    {
        scanf("%d%d%d%d%d%d%d",&shu[0],&shu[1],&shu[2],&shu[3],&shu[4],&shu[5],&n);
        int ans = fun();
        printf("Case %d: %d\n", j, ans);
    }
    return 0;
 }