Hex-a-bonacci ————记忆化搜索_input】input starts with an integer t (t <= 100), d-优快云博客

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
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
Sample Output
Case 1: 216339
Case 2: 79
Case 3: 16636
Case 4: 6
Case 5: 54

题意：

根据这段给定的代码写程序
因为代码是直接递归，所以在 $500ms$ 内很难通过所有的测试样例的，复杂度是 $O\large( n^6 )$
所以我们得去掉重复计算，想到用递推或者记忆化搜索

记忆化搜素：

#include<bits/stdc++.h>
using namespace std;
const int MAXN=1e6+7;
const int mod=10000007;
int f[MAXN];

int fn(int n)
{
    if(n<6)    return f[n];
    if(f[n-1])  return f[n]=f[n-1]%mod+f[n-2]%mod+f[n-3]%mod+f[n-4]%mod+f[n-5]%mod+f[n-6]%mod;
    return  f[n]=fn(n-1)%mod+fn(n-2)%mod+fn(n-3)%mod+fn(n-4)%mod+fn(n-5)%mod+fn(n-6)%mod;
}

int main()
{
    int t,n;
    scanf("%d",&t);
    int T=t;
    while(t--)
    {
        memset(f,0,sizeof(f));
        scanf("%d %d %d %d %d %d %d",&f[0],&f[1],&f[2],&f[3],&f[4],&f[5],&n);
        printf("Case %d: %d\n",T-t,fn(n)%mod);
    }
    return 0;
}

递推：

#include<bits/stdc++.h>
using namespace std;
const int MAXN=1e6+7;
const int mod=10000007;
int f[MAXN];
int main()
{
    int t,n;
    scanf("%d",&t);
    int T=t;
    while(t--)
    {
        scanf("%d %d %d %d %d %d %d",&f[0],&f[1],&f[2],&f[3],&f[4],&f[5],&n);
        for(int i=6;i<=n;i++)
            f[i]=f[i-1]%mod+f[i-2]%mod+f[i-3]%mod+f[i-4]%mod+f[i-5]%mod+f[i-6]%mod;
        printf("Case %d: %d\n",T-t,f[n]%mod);
    }
    return 0;
}