【1006 Hexabonacci（亦是水题）】

Server Time: Sat Aug 11, 2012 2:16 pm

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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

---

PROBLEM SETTER: JANE ALAM JAN

Developed and Maintained by  JANE ALAM JAN

Copyright © 2012  LightOJ, Jane Alam Jan

 

 

 

==========================================================================================================================================

第一便是WA，两个没搞定，，一个是MOD，还有竟然从0case开始了。。。

 

罪过。。

直接上AC代码了。

 

 

// Project name : 1006 ( Hex-a-bonacci ) 
// File name    : main.cpp
// Author       : ismdeep
// E-mail       : ismdeep@live.com
// Date & Time  : Sat Aug 11 14:04:28 2012


#include <iostream>
#include <stdio.h>
#include <string>
#include <cmath>
#include <algorithm>
using namespace std;

/*************************************************************************************/
/* data */
#ifndef MAXN
#define MAXN 10010
#endif

#ifndef MOD
#define MOD 10000007
#endif

typedef unsigned long long longint;

longint _map_[MAXN];

int n;

/*************************************************************************************/
/* procedure */

void init()
{
    for (int i = 0; i < 6; i++)
    {
        cin >> _map_[i];
        _map_[i] %= MOD;
    }
    cin >> n;
}

void cal()
{
    for (int i = 6; i <= n; i++)
    {
        //_map_[i] = _map_[i-1] + _map_[i-2] + _map_[i-3] + _map_[i-4] + _map_[i-5] + _map_[i-6];
        _map_[i] = 0;
        for (int j = 1; j <= 6; j++)
        {
            _map_[i] += _map_[i-j];
            _map_[i] %= MOD;
        }
    }
}

void show_result()
{
    cout << _map_[n] << endl;
}
/*************************************************************************************/
/* main */
int main()
{
    int t;
    cin >> t;
    for (int case_index = 1; case_index <= t; case_index++)
    {
        cout << "Case " << case_index << ": ";
        init      ();
        cal       ();
        show_result();
    }
    return 0;
}