| 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 |
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 10002
using namespace std;
int a[N];
int main()
{
int i,n,caseno=0,cases;
scanf("%d", &cases);
while( cases-- )
{
scanf("%d %d %d %d %d %d %d",&a[0],&a[1],&a[2],&a[3],&a[4],&a[5],&n);
for(i=0;i<=5;i++)
a[i]=a[i]%10000007;
if(n>5)
for(i=6;i<=n;i++)
a[i]=(a[i-1]+a[i-2]+a[i-3]+a[i-4]+a[i-5]+a[i-6])%10000007;开始这忘了、、
printf("Case %d: %d\n",++caseno,a[n]);
}
return 0;
}