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
", ++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
打表缩短时间
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int fn[100001];
int a, b, c, d, e, f;
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);
fn[0]=a;
fn[1]=b;
fn[2]= c;
fn[3]= d;
fn[4]=e;
fn[5]= f;
for(int i=6;i<=n;i++)
{
fn[i]=fn[i-1]% 10000007 + fn[i-2]% 10000007 + fn[i-3]% 10000007 + fn[i-4]% 10000007 + fn[i-5] % 10000007+ fn[i-6]% 10000007 ;
}
printf("Case %d: %d
", ++caseno, fn[n] % 10000007);
}
return 0;
}