Light OJ 1006( 递归优化 )

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

 

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PROBLEM SETTER: JANE ALAM JAN

题意：刚开始还有点懵了，不知道这题要干啥，优化题目中的递归代码。用递归显然会TLE，所以我们得用递推来尝试着写。

#include<iostream>#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>typedef long long ll;using namespace std;#define INF 1e18#define N 10010ll sum[N];int main(){  int T,cas=0;  scanf("%d",&T);  while(T--)  {    int a,b,c,d,e,f,n;    scanf("%d%d%d%d%d%d%d",&a,&b,&c,&d,&e,&f,&n);    memset(sum,0,sizeof(sum));    sum[0]=a,sum[1]=b,sum[2]=c;    sum[3]=d,sum[4]=e,sum[5]=f;    for(int i=6;i<=n;i++)    {      sum[i]=(sum[i-1]+sum[i-2]+sum[i-3]+sum[i-4]+sum[i-5]+sum[i-6])%10000007;    }    printf("Case %d: %lld\n",++cas,sum[n]%10000007);   ///用lld，别用I64d  }  return 0;}