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
#include <stdio.h> #include <string.h> int a, b, c, d, e, f; int m[10005]; int main() { int n, caseno = 0, cases; scanf("%d", &cases); int flag=1; 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; }
原文：http://www.cnblogs.com/JT-L/p/3710269.html