【light-oj】-1006 - Hex-a-bonacci(思维)
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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
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define LL long longint a[10010];int main(){int u,ca=1;scanf("%d",&u);while(u--){for(int i=0;i<6;i++){scanf("%d",&a[i]);a[i]=a[i]%10000007;}int n;scanf("%d",&n);for(int i=6;i<=n;i++){a[i]=a[i-1]+a[i-2]+a[i-3]+a[i-4]+a[i-5]+a[i-6];a[i]=a[i]%10000007;}printf("Case %d: %d\n",ca++,a[n]);}return 0;}
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