lightoj -1006 水题

  

Description

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

把程序贴上去会发现超时，则想办法把复杂问题简单化，附上程序

#include <iostream>#include <stdio.h>#include <cstring>using namespace std;int a, b, c, d, e, f;int m[10010];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);        m[0]=a;        m[1]=b;        m[2]=c;        m[3]=d;        m[4]=e;        m[5]=f;        for(int i=6;i<=n;i++)        {            m[i]=m[i-1]%10000007+m[i-2]%10000007+m[i-3]%10000007                 +m[i-4]%10000007+m[i-5]%10000007+m[i-6]%10000007;        }         printf("Case %d: %d\n", ++caseno, m[n]%10000007);    }    return 0;}