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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PDF (English) | Statistics | Forum |
| 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：
1 #include<stdio.h>
2 #include<string.h>
3 #define H 10000007
4 int shu[10100], n;
5 int fun()
6 {
7 for(int i = 0; i < 6; i++)
8 shu[i] %= H;
9 if(n < 6)
10 return shu[n];
11 for(int i = 6; i <= n; i++)
12 shu[i]=(shu[i-6]+shu[i-5]+shu[i-4]+shu[i-3]+shu[i-2]+shu[i-1])%H;
13 return shu[n];
14 }
15 int main()
16 {
17 int t;scanf("%d",&t);
18 for(int j = 1; j <= t; j++)
19 {
20 scanf("%d%d%d%d%d%d%d",&shu[0],&shu[1],&shu[2],&shu[3],&shu[4],&shu[5],&n);
21 int ans = fun();
22 printf("Case %d: %d\n", j, ans);
23 }
24 return 0;
25 }