Light OJ：1006 Hex-a-bonacci（水题）

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

 

PROBLEM SETTER: JANE ALAM JAN

题目大意：求所给程序的答案。
解题思路：输入a[0]到a[5]，那么计算a[6]的时候就用到了a[5]、a[4]、a[3]、a【2】、a【1】、a【0】，接着计算a【7】就用到了a【6】~【1】。
代码如下：
#include<cstdio>
#define MOD 10000007
int a[10010];
int main()
{
int t;
scanf("%d",&t);
int kcase=1;
while(t--)
{
for(int i=0;i<6;i++)
{
scanf("%d",&a[i]);
a[i]=a[i]%MOD;
}
int n;
scanf("%d",&n);
for(int i=6;i<=n;i++)//一直递推到n
{
a[i]=a[i-1]+a[i-2]+a[i-3]+a[i-4]+a[i-5]+a[i-6];//用到了前六个
a[i]=a[i]%MOD;
}
printf("Case %d: %d\n",kcase++,a
);
}
return 0;
}