lightoj-1006-Hex-a-bonacci【思维】

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1006 - Hex-a-bonacci

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.

题解：用递归会TLE，而且数据会爆 int 

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long
using namespace std;
const LL Q=10000007;
LL a,b,c,d,e,f,n;
LL s[1000010];
int get()
{
  s[0]=a; s[1]=b; s[2]=c;
  s[3]=d; s[4]=e; s[5]=f;
  for(int i=6;i<=n;i++)
    s[i]=(s[i-1]+s[i-2]+s[i-3]+s[i-4]+s[i-5]+s[i-6])%Q;
  return s[n];
}
int main()
{
  int t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%lld %lld %lld %lld %lld %lld %lld",&a,&b,&c,&d,&e,&f,&n);
    printf("Case %d: %lld\n",++text,get()%Q); // 因为 a，b，c，d，e，f可能会大于 Q 
  }
  return 0;
}